3 Marks

Question 513 Marks

An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the sum of the numbers obtained is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered 2, 3, 4, ....., 12 is picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8?

Answer

Let E₁, E₂ and A be the events as defined below:
E₁ = The coin shows a head
E₂ = The coin shows a head
A = The noted number is 7 or 8
$\therefore\ \text{P}(\text{E})_1=\frac{1}{2}$
$\text{P}(\text{E})_2=\frac{1}{2}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{11}{36}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{11}$
Using the law of total probability, we get
Required probability $=\text{P(A)}=\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{1}{2}\times\frac{11}{36}+\frac{1}{2}\times\frac{2}{11}$
$=\frac{11}{72}+\frac{1}{11}$
$=\frac{121+72}{792}$
$=\frac{193}{792}$

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Question 523 Marks

A die is thrown three times. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$, if
A = 4 appears on the third toss,
B = 6 and 5 appear respectively on first two tosses.

Answer

Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and and 5 on second throw
Clearly,
A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4), (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4), ...... (6, 1, 4), (6, 2 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
Now,
$(\text{A}\cap\text{B})=\{(6, 5, 4)\}$
$\therefore\ \text{Required probability} = \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{6}$
$\therefore\ \text{Required probability} = \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{A})}=\frac{1}{36}$

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Question 533 Marks

A coin is tossed three times, if head occurs on first two tosses, find the probability of getting head on third toss.

Answer

Consider the given events.
A = Getting head on third toss
B = Getting head on first two tosses
Clearly,
A = {(H, H, H), (H, T, H), (T, H, H), (T, T, H)}
B = {(H, H, H), (H, H, T)}
Now,
$\text{A}\cap\text{B}=\{\text{H},\text{H},\text{H}\}$
$\therefore\text{Required probability}=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{2}$

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Question 543 Marks

There are 3 red and 5 black balls in bag 'A'; and 2 red and 3 black balls in bag 'B'. One ball is drawn from bag 'A' and two from bag 'B'. Find the probability that out of the 3 balls drawn one is red and 2 are black.

Answer

It si givem that bag A contains 3 red and 5 balck balls (3R, 5B) and bag B contains 2 red and 3 black balls (2R, 3B).
Now,
P(One red and 2 black) = P(one red from bag A and two black from bag B) + P(black ball from bag A and remaining balls from bag B)
$=\frac{3}{8}\times\frac{3}{5}\times\frac{2}{4}+\frac{5}{8}\times\frac{2}{5}\times\frac{3}{4}\times2$
$=\frac{9}{80}+\frac{30}{80}$
$=\frac{39}{80}$
Note: 2 is multiplied by second term because there are two ways to select red and black balls from bag B.
While the first way is to pick black ball first, followed by red, the second way is to pick red ball first, followed by black.

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Question 553 Marks

A pair of dice is thrown. Find the probability of getting 7 as the sum if it is known that the second die always exhibits a prime number.

Answer

Consider the given events
A = A prime number appears on second die.
B = The sum of the numbers on two dice is 7.
Clearly,

A = {(1, 2), (1, 3), (1, 5),

(2, 2), (2, 3), (2, 5),

(3, 2), (3, 3), (3, 5),

(4, 2), (4, 3), (4, 5),

(5, 2), (5, 3), (5, 5),

(6, 2), (6, 3), (6, 5)}

B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}
Now,
$\text{A}\cap\text{B}=\{(2, 5), (5, 2), (4, 3)\}$
$\therefore\ \text{Required probability}=\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{A})}=\frac{3}{18}=\frac{1}{6}$

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Question 563 Marks

Three machines E₁, E₂, E₃ in a certain factory produce 50%, 25% and 25%, respectively, of the total daily output of electric bulbs. It is known that 4% of the tubes produced one each of the machines E₁and E₂ are defective, and that 5% of those produced on E₃ are defective. If one tube is picked up at random from a day's production, then calculate the probability that it is defective.

Answer

Let D be the event that the picked up tube is defective.
Let A₁, A₂ and A₃ be the events that the tube is produced on macjines E₁, E₂ and E₃ respectively.
P(D) = P(A₁) P(D|A₁) + P(A₂) P(D|A₂) + P(A₃) P(D|A₃) .....(i)
$\text{P}(\text{A}_1)=\frac{50}{100}=\frac{1}{2},\text{P}(\text{A}_2)=\frac{25}{100}=\frac{1}{4},\text{P}(\text{A}_3)=\frac{25}{100}=\frac{1}{4}$
$\text{P}(\text{D}|\text{A}_1)=\text{P}(\text{D}|\text{A}_2)=\frac{4}{100}=\frac{1}{25}$
$\text{P}(\text{D}|\text{A}_3)=\frac{5}{100}=\frac{1}{20}$
Putting these values in (i), we get
$\text{P(D)}=\frac{1}{2}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{20}$
$\text{P(D)}=\frac{17}{400}$

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Question 573 Marks

Three cards are drawn with replacement from a well shuffled pack of 52 cards. Find the probability that the cards are a king, a queen and a jack.

Answer

Three cards are drawn with replacement fron a pack of cards
There are 4 Kings, 4 Queens, 5 Jacks.
P (1 King, 1 Queen, 1 Jack)
$=\text{P}\big[(\text{K}\cap\text{Q}\cap\text{J})\cup(\text{K}\cap\text{J}\cap\text{Q})\cup(\text{J}\cap\text{K}\cap\text{Q})\\\cup(\text{J}\cap\text{Q}\cap\text{K})\cup(\text{Q}\cap\text{K}\cap\text{L})\cup(\text{Q}\cap\text{J}\cap\text{K})\big]$
$=\text{P}(\text{K}\cap\text{Q}\cap\text{J})+\text{P}(\text{K}\cap\text{J}\cap\text{Q})+\text{P}(\text{J}\cap\text{K}\cap\text{Q})\\+\text{P}(\text{J}\cap\text{Q}\cap\text{K})+\text{P}(\text{Q}\cap\text{K}\cap\text{L})+\text{P}(\text{Q}\cap\text{J}\cap\text{K})$
$=\text{P}(\text{K}) \text{P}(\text{Q}) \text{P}(\text{J})+\text{P}(\text{K}) \text{P}(\text{J}) \text{P}(\text{Q})+\text{P}(\text{J}) \text{P}(\text{K}) \text{P}(\text{Q}) \\ +\text{P}(\text{J}) \text{P}(\text{Q}) \text{P}(\text{K})+\text{P}(\text{Q}) \text{P}(\text{K}) \text{P}(\text{L})+\text{P}(\text{Q}) \text{P}(\text{J}) (\text{K})$
$=\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52} \\ +\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}$
$=\frac{6}{13\times13\times13}$
$=\frac{6}{2197}$
Required probability $=\frac{6}{2197}$

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Question 583 Marks

If a machine is correctly set up it produces 90% acceptable items. If it is incorrectly set up it produces only 40% acceptable item. Past experience shows that 80% of the setups are correctly done. If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly set up.

Answer

Let A be the event that the machine produces two acceptable items.
Also, let E₁ represent the event that the machine is correctly set up and E₂ represent the event that the machine is incorrectly set up
$\therefore\ \text{P}(\text{E}_1)=0.8$
$\text{P}(\text{E}_2)=0.2$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=0.9\times0.9=0.81$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=0.40\times0.40=0.16$
Using Bayes, theorem, we get
Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{0.8\times0.81}{0.8\times0.81+0.2\times0.61}$
$=\frac{81}{85}$

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Question 593 Marks

A bag contains 1 white and 6 red balls, and a second bag contains 4 white and 3 red balls. One of the bags is picked up at random and a ball is randomly drawn from it, and is found to be white in colour. Find the probability that the drawn ball was from the first bag.

Answer

Let A, E₁ and E₂ denote the events that the ball is white, bag I is chosen and bag II is chosen, respectively.
$\therefore\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{1}{7}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{4}{7}$
Using Baye's therorem, we get
Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\times\frac{1}{7}}{\frac{1}{2}\times\frac{1}{7}+\frac{1}{2}\times\frac{4}{7}}$
$=\frac{1}{1+4}=\frac{1}{5}$

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Question 603 Marks

If A and B are two independent events such that $\text{P}(\text{A}\cap\text{B})=0.60$ and P(A) = 0.2, find P(B).

Answer

$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A})\times(\text{B})$
[$\because$ A and B are independent events]
$\Rightarrow\ 0.6=0.2+\text{P(B)}=0.2\times\text{P(B)}$
$\Rightarrow\ 0.6-0.2=\text{P(B)}(1-0.2)$
$\Rightarrow\ \text{P(B)}=\frac{0.6-0.2}{1-0.2}$
$\Rightarrow\ \text{P(B)}=\frac{0.4}{0.8}$
$\Rightarrow\ \text{P(B)}=\frac{1}{2}$
$\Rightarrow\ \text{P(B)}=0.5$

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Question 613 Marks

A, B, and C are independent witness of an event which is known to have occurred. A speaks the truth three times out of four, B four times out of five and C five times out of six. What is the probability that the occurrence will be reported truthfully by majority of three witnesses?

Answer

P(A speaks truth) $=\frac{3}{4}$
P(B speaks truth) $=\frac{4}{5}$
P(C speaks truth) $=\frac{5}{6}$
P(majority speaks truth) = P(two speaks truth) + P(all speak truth)
= P(A) × P(B)[1 - P(C)] + P(A) × P(C)[1 - P(B)] + P(C) × P(B)[1 - P(A)] + P(A) × P(B) × P(C)
$=\frac{3}{4}\times\frac{4}{5}\Big(1-\frac{5}{6}\Big)+\frac{3}{4}\times\frac{5}{6}\Big(1-\frac{4}{5}\Big) \\ +\frac{4}{5}\times\frac{5}{6}\Big(1-\frac{3}{4}\Big)+\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}$
$=\frac{12}{120}+\frac{15}{120}+\frac{20}{120}+\frac{60}{120}$
$=\frac{107}{120}$

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Maths 3 Marks