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PROBABILITY · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 91 questions.

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Question 11 Mark
If $\frac2{11}$ is the probability of an event A, what is the probability of the event 'not A'.
Answer
Given, probability of event A i.e., P(A) = $\frac2{11}$
$\therefore$ Probability ('not A')= $P(\bar A)$
$=1-P(A)=1-\frac2{11}=\frac9{11}$
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Question 21 Mark
Three coins are tossed once. Find the probability of getting: almost two tails
Answer
When three coins are tossed then the sample space S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting at most 2 tails
n(A) = 7
P(getting almost 2 tails) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{7}{\mathrm{8}}$
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Question 31 Mark
Three coins are tossed once. Find the probability of getting: no tail
Answer
When three coins are tossed then Sample space S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
And thus here n(S) = 8
Let A be the event of getting no tails
n(A) = 1
P(getting no tails) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{1}{\mathrm{8}}$
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Question 41 Mark
Three coins are tossed once. Find the probability of getting: exactly two tails
Answer
When three coins are tossed then total outcomes S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting exactly 2 tails
n(A) = 3
P(getting exactly 2 tails) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3}{\mathrm{8}}$
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Question 51 Mark
Three coins are tossed once. Find the probability of getting: 3 tails
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting 3 tails
n(A) = 1
P(getting 3 tails) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{1}{\mathrm{8}}$
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Question 61 Mark
Three coins are tossed once. Find the probability of getting: no head
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting no heads
n(A) = 1
P(getting no head) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{1}{\mathrm{8}}$
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Question 71 Mark
Three coins are tossed once. Find the probability of getting: at most 2 heads
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
let A be the event of getting at most 2 heads
n(A) = 7
P(getting at most 2 heads) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{7}{8}$
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Question 81 Mark
Three coins are tossed once. Find the probability of getting: at least 2 heads
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting at least 2 head
n(A) = 4
P(getting atleast 2 heads) = P(A) = $\frac{n(A)}{n(S)}=\frac{4}{8}=\frac{1}{2}$
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Question 91 Mark
Three coins are tossed once. Find the probability of getting: 2 heads
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting 2 heads
n(A) = 3
P(getting 2 heads) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3}{\mathrm{8}}$
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Question 101 Mark
Three coins are tossed once. Find the probability of getting: 3 heads
Answer
When three coins are tossed then total outcomes, S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}
Where s is sample space and here n(S) = 8
Let A be the event of getting 3 heads n(A) = 1
P(getting 3 heads) = P(A) = $\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{1}{\mathrm{8}}$
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Question 111 Mark
A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is 12.
Answer
The coin with 1 marked on one face and 6 on the other face.
The coin and die are tossed together.
$\therefore$ S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
$\Rightarrow$ smpale space n(S) = 12
Let B be the event having a sum of the numbers is 12
therefore, B = {(6, 6)}
$\Rightarrow$ n (B) = 1
Thus, $P(B) = \frac{1}{{12}}$
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Question 121 Mark
A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is 3.
Answer
The coin with 1 marked on one face and 6 on the other face.
The coin and die are tossed together.
$\therefore$ S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
$\Rightarrow$ sample space of event is given by n(S) = 12
Let A be the event having sum of numbers is 3
$\therefore$ A = {(1, 2)}
$\Rightarrow$ n (A) = 1
Thus, P(A) = $\frac{1}{{12}}$
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Question 131 Mark
A card is selected from a pack of 52 cards. Calculate the probability that the card is black card.
Answer
Let C be the event of drawing a black card. Now, we know that in a pack of 52 cards, there are 26 black cards.
$\therefore P(C) =\frac{{number of favourable outcomes}}{{number of total outcomes}}= \frac{{26}}{{52}} = \frac{1}{2}$
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Question 141 Mark
A card is selected from a pack of 52 cards. Calculate the probability that the card is an ace.
Answer
Let B be the event of drawing an ace. We know that in a pack of 52 card there are four aces.
$\therefore P(B) = \frac{4}{{52}} = \frac{1}{{13}}$
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Question 151 Mark
A card is selected from a pack of 52 cards. Calculate the probability that the card is an ace of spades.
Answer
Let A be the event of drawing an ace of spades. Now there is only once ace of spade.
Therefore,
$ P(A) = \frac{1}{{52}}$
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Question 161 Mark
A card is selected from a pack of 52 cards. How many points are there in the sample space?
Answer
We have to draw One card from a pack of 52 cards.
$\Rightarrow$ Number of points in the sample space S = n(S) = 52.
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Question 171 Mark
A die is thrown, find the probability of A number less than 6 will appear.
Answer
Here the sample space of the event is given by
S = {1, 2, 3, 4, 5, 6}
$\therefore$ n(S) = 6
Let E be the event of getting a number less than 6
E = {1, 2, 3, 4, 5} $\Rightarrow$ n(E) = 5
Thus, $P(E) = \frac{{n(E)}}{{n(S)}} = \frac{5}{6}$
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Question 181 Mark
A die is thrown. Find the probability of getting a number more than 6.
Answer
We have to find the probability of getting a number of more than 6
The sample space associated with the random experiment of rolling a die is given by
S = {1, 2, 3, 4 ,5 , 6}. Clearly, there are 6 elements in S.
$\therefore$The total number of elementary events = 6.
Since no face of the die is marked with a number greater than 6.
So, favourable number of elementary events = 0
Hence, required probability = $\frac{0}{6}$= 0
In fact, the given event is an impossible event. So, the probability of its occurrence is zero.
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Question 191 Mark
A die is thrown, find the probability of A number less than or equal to 1 will appear.
Answer
In a throw of a die, sample space S = {1, 2, 3, 4, 5, 6}
$\therefore$ n(S) = 6
Let C be the event of getting a number less than or equal to 1
Elements of C = {1} $\Rightarrow$ n(C) = 1
Thus $P(C)=\frac{{n(C)}}{{n(S)}} = \frac{1}{6}$
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Question 201 Mark
A die is thrown. Find the probability of getting a number greater than or equal to 3.
Answer
We have to find the probability of getting a number greater than or equal to 3.
The sample space associated with the random experiment of rolling a die is given by
S = {1, 2, 3, 4 ,5 , 6}. Clearly, there are 6 elements in S.
$\therefore$ The total number of elementary events = 6.
A number greater than or equal to 3 is obtained, if we get any one of 3, 4,5 , 6 as an outcome.
So, favourable number of elementary events = 4
Hence, the required probability =$\frac{4}{6}=\frac{2}{3}$
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Question 211 Mark
A die is thrown. Find the probability that a prime number will appear.
Answer
As 2, 3, 5 are prime numbers up to 6, so the desired outcomes are 2, 3, 5, and total outcomes are 1, 2, 3, 4, 5, 6
Therefore, total no. of outcomes are 6, and total no. of desired outcomes are 3
Probability of getting a prime number = $\frac{3}{6}$ = $\frac{1}{2}$
Conclusion: Probability of getting a prime number when a die is thrown is $\frac{1}{2}$
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Question 221 Mark
A coin is tossed twice, what is the probability that atleast one tail occurs?
Answer
Since a coin tossed twice,
so the sample space (S) is given by S= {HH, HT, TH, TT}
$\therefore$ Total number of possible out comes n (S) = 4
Let E be the event of getting at least one tail
$\therefore$ n(E) = 3
$\therefore$ Probability of getting at least one tail ${P(E)=\frac{n(E)}{n(S)}=\frac34}$
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Question 231 Mark
Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = $\{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$
Assignment $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
  $\frac{1}{14}$ $\frac{2}{14}$ $\frac{3}{14}$ $\frac{4}{14}$ $\frac{5}{14}$ $\frac{6}{14}$ $\frac{15}{14}$
Answer
The conditions of axiomatic approach in the given assignment are being fulfilled as $p(w_7)=\frac{15}{14}$>1 and probability should be less than or equal to one and positive.
Hence, the given assignment is not valid.
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Question 241 Mark
Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = $\{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$
Assignment $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
  -0.1 0.2 0.3 0.4 -0.2 0.1 0.3
Answer
The conditions of axiomatic approach do not hold true in the given assignment, because $p(w_1)$ and $p(w_4)$ is negative.
To be the conditions true each of the number $p(w_i)$ should be less than or equal to one and positive.
So, the assignment is not valid
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Question 271 Mark
P(A) P(B) P(A$\cap$B) P(A$\cup$B)
$\frac 13$ $\frac 15$ $\frac 1{15}$ _____
Answer
We know that
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$
Putting Values
$P(A\cup B) = \frac 13 + \frac 15 - \frac 1{15}$
$P(A\cup B) = \frac {5\; +\; 3\; -\; 1}{15}$
$P(A\cup B) = \frac 7{15}$
Hence $P(A\cup B) = \frac 7{15}$
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Question 281 Mark
Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = $\{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$
Assignment $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
  0.1 0.2 0.3 0.4 0.5 0.6 0.7
Answer
Both the conditions of axiomatic approach in the given assignment are
  1. Each of the number $p(w_i)$ is less than or equal to one and is positive,
  2. Sum of probabilities is 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 > 1.
It's clear that the second condition is not satisfied as the sum should be exactly equal to one.
Hence, the given assignment is not valid.
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Question 291 Mark
Check whether the probabilities P(A) and P(B) are consistently defined P(A) = 0.5, P(B) = 0.4, $P(A \cup B) = 0.8$
Answer
Given that P(A) = 0.5, P(B) = 0.4 and $P(A \cup B) = 0.8$
Applying the general addition rule,

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\therefore \;0.8 = 0.5 + 0.4 - P(A \cap B)$
$\Rightarrow \;P(A \cap B) = 0.9 - 0.8 = 0.1$
$\therefore \;P(A \cap B) < P(A)\;{\text{and}}\;P(A \cap B) < P(B)$
Thus the given probabilities are consistently defined.

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Question 301 Mark
In a simultaneous throw of a pair of dice, find the probability of getting an even number on one and a multiple of 3 on the other.
Answer
We have to find the probability of getting an even number on one and a multiple of 3 on the other.
Let E be the event of getting even on one and multiple of three on other.
E = {(2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4) (3,6) (6,2) (6,4)}
n(E) = 11
$P(E)=\frac{n(E)}{n(S)}$
$P(E)=\frac{11}{36}$
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Question 311 Mark
In a simultaneous throw of a pair of dice, find the probability of getting 8 as the sum.
Answer
Since a pair of dice have been thrown,
$\therefore$ Numbers of elementary events in sample space is $6^2 = 36$
Suppose E be the event that the sum 8 appear on the faces of dice,
$\therefore$ E = {(2,6), (3,5), (4,4), (5,3), (6,2)}
$\therefore$ $n(E) = 5$
$\therefore$ $P (E) =$ $\frac{5}{36}$
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Question 321 Mark
Check whether the probabilities P(A) and P(B) are consistently defined P(A) = 0.5. P(B) = 0.7, $P(A \cap B) $ = 0.6
Answer
Given that P(A) = 0.5, P(B) = 0.7 and $P(A \cap B) = 0.6$
Now, as $P(A \cap B) > P(A)$,
We can say that the given probabilities are not consistently defined.
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Question 331 Mark
Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = $\{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$
Assignment $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
  $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$
Answer
Both the conditions of axiomatic approach hold true in the given assignment, that is
  1. Each of the number $p(w_i)$ is less than or equal to one and is positive
  2. Sum of probabilities is $\frac 17 +\frac 17+\frac 17+\frac 17+\frac 17+\frac 17+\frac 17+\frac 77 = 1$
Hence, the given assignment is valid.
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Question 341 Mark
Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = $\{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$
Assignment $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
  0.1 0.01 0.05 0.03 0.01 0.2 0.6
Answer
Both the conditions of axiomatic approach hold true in the given assignment, that is
  1. Each of the probability $p(w_i)$ is less than one and is positive
  2. Sum of probabilities is 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1
Hence,the given assignment is valid.
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Question 351 Mark
Two dice are thrown. The events A, B, and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice $\le$ 5.
State true or false: A′, B′, C are mutually exclusive and exhaustive (give the reason for your answer).
Answer
False.
If two dice are thrown then, total number of possible outcomes S = 6 $\times$ 6 = 36
S = { (1, 1), (1, 2) , (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
A = getting an even number on the first die
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
B = getting an odd on the first die.
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
C = getting the sum of the numbers on the die $\le$ 5
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
So, A′, B′, C are not mutually exclusive and exhaustive
$\because$ A' = B , B' = A
$\therefore$ A' $\cap$ B' $\cap$ C = B $\cap$ A $\cap$ C = $\phi$
but A' $\cap$ C = B $\cap$ C $\ne$ $\phi$ so, A' , B' and C are not mutually exclusive..
A' $\cup$ B' $\cup$ C = B $\cup$ A $\cup$ C = S {sample space}
Hence, A', B', and C are exhaustive.
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Question 361 Mark
Two dice are thrown. The events A, B, and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice $\le$ 5.
State true or false: A and B′ are mutually exclusive. (give the reason for your answer).
Answer
False.
Solution: If two dice are thrown then, total number of possible outcomes S = 6 $\times$ 6 = 36
S = { (1, 1), (1, 2) , (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
A = getting an even number on the first die
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
B = getting an odd on the first die.
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
B' = S - B
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
C = getting the sum of the numbers on the die $\times$ 5
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
False,
$\because$ B' = A
and A $\cup$ A = A $\ne$ $\phi$
Hence, A and B' are not mutually exclusive.
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Question 371 Mark
Two dice are thrown. The events A, B, and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice $\le$ 5.
State true or false: A and C are mutually exclusive (give the reason for your answer).
Answer
False.
Solution: If two dice are thrown then, total number of possible outcomes S = 6 $\times$ 6 = 36
S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
A = getting an even number on the first die
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
B = getting an odd on the first die.
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
C = getting the sum of the numbers on the die $\le$ 5
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
So, the statement that A and C are mutually exclusive is false.
$\because$ A $\cap$ C = {(2, 1), (2, 2), (2, 3), (4, 1)} $\neq \phi$
$\therefore$ A and C are not mutually exclusive.
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Question 381 Mark
Two dice are thrown. The events A, B, and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice $\le$ 5.
State true or false: A = B′ (give the reason for your answer).
Answer
True
Solution: If two dice are thrown then, total number of possible outcomes S = 6 $\times$ 6 = 36
S = { (1, 1), (1, 2) , (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
A = getting an even number on the first die
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
B = getting an odd on the first die.
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
B' = S - B
={(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
C = getting the sum of the numbers on the die $\le$ 5
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
True, A = B'
A = B' { $\because$ A and B are mutually exclusive}.
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Question 391 Mark
Two dice are thrown. The events A, B, and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice $\le$ 5.
state true or false: A and B are mutually exclusive and exhaustive (give the reason for your answer).
Answer
True
Solution: If two dice are thrown then, total number of possible outcomes S = 6 $\times$ 6 = 36
S = { (1, 1), (1, 2) , (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
A = getting an even number on the first die
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
B = getting an odd on the first die.
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
C = getting the sum of the numbers on the die $\le$ 5
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
True, A and B are mutually exclusive and exhaustive
$\because$A $\cup$ B = S {if A union B is equal to whole sample space then, it is exhaustive.}
$\because$ A = getting an even number on the first die.
B = getting an odd number on the first die.
$\therefore$ A $\cap$ B = $\phi$,
$\therefore$ A and B are mutually exclusive and exhaustive.
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Question 401 Mark
Two dice are thrown. The events A, B, and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice $\le$ 5.
State true or false: A and B are mutually exclusive (give the reason for your answer).
Answer
True
Solution: If two dice are thrown then, total number of possible outcomes S = 6 $\times$ 6 = 36
S = { (1, 1), (1, 2) , (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
A = getting an even number on the first die
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
B = getting an odd on the first die.
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
C = getting the sum of the numbers on the die $\le$ 5
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
True, A and B are mutually exclusive
$\because$ A = getting an even number on the first die.
B = getting an odd number on the first die.
A $\cap$ B = $\phi$
$\therefore$ A and B are mutually exclusive events.
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Question 411 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event '$A \cap B'\cap C'$'
Answer
Given that two dice are thrown then sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event A: getting an even number on the first die.
Outcomes of event A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2,6),
(4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
B' = S - B
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2,6),
(4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = A
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
C' = S - C
= {(1, 5), (1, 6),(2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6),(4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Clearly, $A \cap B' \cap C'$ = {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6,6)}
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Question 421 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event B and C
Answer
Given that two dice are thrown then sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event A: getting an even number on the first die.
Outcomes of event A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2,6),
(4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event B: getting an odd number on the first die
Outcomes of event B ={(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
B and C = $B \cap C$ = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}
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Question 431 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\leq 5$
Describe the event 'B or C'
Answer
Given that two dice are thrown then space space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event A: getting an even number on the first die.
Outcomes of event A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2,6),
(4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\leq 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
B or C = $B \cup C$ =
{(1,1), (1,2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (5,1),(5,2), (5, 3), (5, 4), (5, 5), (5, 6)}
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Question 441 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event A but not C
Answer
Given that two dice are thrown then sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event A: getting an even number on the first die.
Outcomes of event A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2,6),
(4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
A but not C = A - C
= {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
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Question 451 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event A and B
Answer
Given that two dice are thrown then sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event A: getting an even number on the first die.
Outcomes of event A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2,6),
(4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
A and B = $A \cap B = \phi$
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Question 461 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event A or B
Answer
Given that two dice are thrown then sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event A: getting an even number on the first die.
Outcomes of event A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2,6),
(4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event B: getting an odd number on the first die
Outcomes of event B ={(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

A or B =$A \cup B$ = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, l), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = S

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Question 471 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event not B
Answer
Given that two dice are thrown. So sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event A: getting an even number on the first die.
Outcomes of event A= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2,6),
(4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C= {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
not B = {(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = A
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Question 481 Mark
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die
B: getting on odd number on the first die
C: getting the sum of the numbers on the dice $\le 5$
Describe the event A'
Answer
Given that two dice are thrown, so sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, l), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event A: getting an even number on the first die.
Outcomes of event A= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2,6),
(4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Event B: getting an odd number on the first die
Outcomes of event B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Event C: getting the sum of the number on the dice $\le 5$
Outcomes of event C = {(1, 1),(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

A' = S - A
= (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} = B

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Question 491 Mark
Three coins are tossed. Describe Three events which are mutually exclusive but not exhaustive.
Answer
Given that three coins are tossed then sample space (S) is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let event A: getting three heads = {HHH},
event B: getting exactly two heads = {HHT, HTH, THH} and
event C: getting three tails = {TTT}
Now $\style{font-size:28px}{A\cap B=\phi}$, $\style{font-size:28px}{B\cap C=\phi}$ and $\style{font-size:28px}{A\cap C=\phi}$
Thus A, B, C are mutually exclusive events
Also ${A\cup B\cup C=\{HHH,\;HHT,\;HTH,\;THH,\;TTT\;\}⧧S}$
Thus A, B, C are mutually exclusive but not exhaustive events.

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Question 501 Mark
Three coins are tossed. Describe Two events which are mutually exclusive but not exhaustive.
Answer
Given that three coins are tossed then sample space (S) is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let event A: getting three heads = {HHH} and
event B: getting three tails = {TTT}
Now ${A\cap B=\phi}$ and ${AUB=\{HHH,\;TTT\;\}⧧S}$
Thus A and B are mutually exclusive but not exhaustive.
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(Each question 1 marks) - MATHS STD 11 Science Questions - Vidyadip