5 Marks Questions

Question 15 Marks

Three coins are tossed simultaneously. Consider the event $E$ 'three heads or three tails,' $F$ 'at least two heads' and $G$ 'at most two heads'. Of the pairs $(E, F),(E, G)$ and $(F, G)$, which are independent ? Which are dependent?

Answer

The sample space of the experiment is
$S=\{HHH, HHT, HTH, THH, HTT, THT, TTH,TTT\} $
$\therefore \quad n(S)=8$
Here,
$\begin{aligned} & E=\{ HHH , TTT \} ; n(E)=2 \\ & F=\{ HHH , HHT , HTH , THH \}, n(F)=4\end{aligned}$
and
$
\begin{aligned}
G &=\{HHT, HTH, THH, HTT, THT, TTH, TTT\} \\
n(G) & =7
\end{aligned}
$
Also,
$\begin{aligned} E \cap F & =\{ HHH \}, n(E \cap F)=1 \\ E \cap G & =\{ TTT \}, n(E \cap G)=1 \\ F \cap G & =\{ HHT , HTH , THH \}, n(F \cap G)=3\end{aligned}$
$
\begin{aligned}
\text { Therefore, } \quad\quad P(E) & =\frac{n(E)}{n(S)}=\frac{2}{8}=\frac{1}{4} \\
P(F) & =\frac{n(F)}{n(S)}=\frac{4}{8}=\frac{1}{2} \\
P(G) & =\frac{n(G)}{n(S)}=\frac{7}{8} \\
\text { and } \quad\quad\quad P(E \cap F) & =\frac{n(E \cap F)}{n(S)}=\frac{1}{8} \\
P(E \cap G) & =\frac{n(E \cap G)}{n(S)}=\frac{1}{8} \\
P(F \cap G) & =\frac{n(F \cap G)}{n(S)}=\frac{3}{8} \\
\text { Also, } \quad P(E) \cdot P(F) & =\frac{1}{4} \times \frac{1}{2}=\frac{1}{8} \\
P(E).P(G) & =\frac{1}{4} \times \frac{7}{8}=\frac{7}{32} \\
\text { and } \quad\quad P(f).P(G) & =\frac{1}{2} \times \frac{7}{8}=\frac{7}{16} \\
\end{aligned}
$
Thus,
$
\begin{array}{l}
P(E \cap F)=P(E) \cdot P(F) \\
P(E \cap G) \neq P(E) \cdot P(G) \\
P(F \cap G) \neq P(F) \cdot P(G)
\end{array}
$
Hence, the events ( $E$ and $F$ ) are independent, and the events $(E$ and $G)$ and $(F$ and $G)$ are dependent.

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Question 25 Marks

In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random. Find the probability that
(i) The student opted for NCC or NSS.
(ii) The student has opted for neither NCC nor NSS.
(iii) The student has opted for NSS but not NCC.

Answer

Let $C$ and $S$ denote the event that the student opted for NCC and NSS respectively.
Then,
$
\begin{aligned}
P(C) & =\frac{30}{60}=\frac{1}{2} \\
P(S) & =\frac{32}{60}=\frac{8}{15} \\
P(C \cap S) & =\frac{24}{60}=\frac{2}{5}
\end{aligned}
$
(i) We know :
$\begin{aligned} P(C \cup S) & =P(C)+P(S)-P(C \cap S) \\ & =\frac{1}{2}+\frac{8}{15}-\frac{2}{5} \\ & =\frac{15+16}{30}-\frac{2}{5} \\ & =\frac{31}{30}-\frac{2}{5} \\ & =\frac{19}{30} .\end{aligned}$
$\therefore$ The probability that the student opted for NCC or NSS is $\frac{19}{30}$.
(ii)
$\begin{aligned} P\left(C^{\prime} \cap S^{\prime}\right)=P(C \cup S)^{\prime} & =1-P(C \cup S) \\ & =1-\frac{19}{30}=\frac{11}{30} .\end{aligned}$
$\therefore$ The probability that the student opted neither for NCC nor NSS is $\frac{11}{30}$.
(iii)
$
\begin{aligned}
P(S-C) & =P(S)-P(C \cap S) \\
& =\frac{8}{15}-\frac{2}{5}=\frac{8-6}{15}=\frac{2}{15} .
\end{aligned}
$
$\therefore$ The probability that the students opted for NSS but not NCC is $2 / 15$.

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Question 35 Marks

A fair coin is tossed four times, and a person win ₹ 1 for each head and lose ₹ 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Answer

(i) When no head and 4 tails appear. Let A be the event money lost $=₹ (4 \times 1.50)=₹ 6.00$.
There is only one way of getting no head and 4 tails i.e., $($ TTTT $) \Rightarrow n(A)=1$
$n(S)=16$, since there are 16 possible outcomes
$\therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{1}{16}$
(ii) Let $B$ be the event when 1 head and 3 tails appear.
$\therefore \quad B=\{\text { HTTT, THTT, TTHT, TTTH }\}$
$\Rightarrow \quad n(B)=4$
Money lost $=₹ (3 \times 1.50-1 \times 1)=₹ 3.50$
$\therefore \quad P(B)=\frac{n(B)}{n(S)}=\frac{4}{16}=\frac{1}{4}$
(iii) Let C be the event that 2 head and 2 tail appear.
$\therefore \quad$ Money lost $=₹ (2 \times 1.50-2 \times 1)$
=₹ 1
$\begin{array}{r}C=\{\text { HHTT, HTHT, HTTH, THHT, THTH, TTHH }\}\end{array}$
$\begin{array}{ll}\Rightarrow & n(C)=6 \\ \therefore & P(C)=\frac{n(C)}{n(S)}=\frac{6}{16}=\frac{3}{8}\end{array}$
(iv) Let D be the event that 3 head and 1 tail appear.
$\therefore \quad D=\{$ HHHT, HHHT, THHH, HHTH $\}$
$\Rightarrow \quad n(D)=4$
Money gained $=₹ (3 \times 1-1 \times 1.5)=₹ 1.50$
$P(E)=\frac{n(D)}{n(S)}=\frac{4}{16}=\frac{1}{4}$
(v) Let E be the event that all heads appear:
$\therefore \quad E=\{HHHH\} \Rightarrow n(E)=1$
Money gained $=₹ (4 \times 1)=₹ 4$
Also, $\quad P(E)=\frac{n(E)}{n(S)}=\frac{1}{16}$.

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Question 45 Marks

If the letters of the word 'ASSASSINATION' are arranged at random. Find the probability that
(i) four $S$ 's come consecutively in the word.
(ii) two I's and two N's come together.
(iii) all $A$ 's are not coming together.

Answer

Total number of letters in the word 'ASSASSINA--TION' are 13.
Out of which 3 A's, 4 S's, 2 I's, 2N's, 1 T's and 1 O's.
(i) If four S's come consecutively in the word, then we considered these 4 S's as 1 group. Now, the number of letters is 10 .

$\text { Number of words all S's are together }=\frac{10!}{3!2!2!}$
Total number of word using letters of the word 'ASSASSINATION'
$=\frac{13!}{3!4!2!2!}$
$\begin{aligned} \therefore \text { Required probability } & =\frac{10!}{3!2!2!} \times \frac{3!4!2!2!}{13!} \\ & =\frac{10!\times 4!}{13!}=\frac{4!}{13 \times 12 \times 11} \\ & =\frac{24}{1716}=\frac{2}{143} .\end{aligned}$
(ii) If 2 I 's and 2 N 's come together, then there as 10 alphabets.
Number of word when 2 I's and 2 N 's are come together
$=\frac{10!}{3!4!} \times \frac{4!}{2!2!}$
$\begin{aligned} \therefore \text { Required probability } & =\frac{\frac{10!4!}{3!4!2!2!}}{\frac{13!}{3!4!2!2!}} \\ & =\frac{4!10!}{2!2!3!4!} \times \frac{3!4!2!2!}{13!} \\ & =\frac{4!10!}{13!}=\frac{4!}{13 \times 12 \times 11} \\ & =\frac{24}{13 \times 12 \times 11}=\frac{2}{143} .\end{aligned}$
(iii) If all A's are coming together, then there are 11 alphabets. Number of words when all A's come together
$=\frac{11!}{4!2!2!}$
Probability when all A's come together
$
\begin{array}{l}
=\frac{\frac{11!}{4!2!2!}}{\frac{13!}{4!3!2!2!}} \\
=\frac{11!}{4!2!2!} \times \frac{4!3!2!2!}{13!} \\
=\frac{11!\times 3!}{13!}=\frac{6}{13 \times 12}=\frac{1}{26}
\end{array}
$
Required probability when all A's does not come together
$=1-\frac{1}{26}=\frac{25}{26} .$

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Question 55 Marks

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that
(i) all the three balls are white
(ii) all the three balls are red.
(iii) One ball is red and two balls are white.

Answer

$\because$ Number of red balls $=5$
and number of white balls $=5$
(i) $P$ (all the three balls are white)
$
\begin{array}{l}
=\frac{{ }^5 C_3}{{ }^{13} C_3}=\left(\frac{\frac{5 \times 4}{2}}{\frac{13 \times 12 \times 11}{3 \times 2}}\right) \\
=\frac{5 \times 4 \times 3}{13 \times 12 \times 11}=\frac{5}{143}
\end{array}
$
(ii) $P$ (all the three balls are red)
$
\begin{array}{l}
=\frac{{ }^8 C_3}{{ }^{13} C_3}=\left(\frac{\frac{8 \times 7 \times 6}{3 \times 2 \times 1}}{\frac{13 \times 12 \times 11}{3 \times 2 \times 1}}\right) \\
=\frac{8 \times 7 \times 6}{13 \times 12 \times 11}=\frac{28}{143}
\end{array}
$
(iii) P (one ball is red and two balls are white)
$
\begin{array}{l}
=\frac{{ }^8 C_1 \times{ }^5 C_2}{{ }^{13} C_3} \\
=\frac{8 \times 10}{13 \times 12 \times 11}=\frac{40}{143} .
\end{array}
$

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Question 65 Marks

Two dice are thrown together. What is the probability that sum of the numbers on the two faces is neither divisible by 3 nor 4 ?

Answer

Let ' $S$ ' be the sample space.
$
\begin{aligned}
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)\} \\
\Rightarrow \quad & n(S)=36
\end{aligned}
$
Let ' $A$ ' and ' $B$ ' be the events that the sum of the numbers on the two faces is divisible by 3 and 4 respectively.
$\begin{array}{l} \therefore A=\{(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3),(6,6)\} \\ \Rightarrow \quad n(A)=12\end{array}$
and, $\quad B=\{(1,3),(2,2),(2,6),(3,5)$,$(4,4),(5,3),(6,2),(3,1),(6,6)\}$
$\Rightarrow \quad n(B)=9$
$\therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{12}{36}$
$P(B)=\frac{n(B)}{n(S)}=\frac{9}{36}$
$A \cap B=\{(6,6)\} \Rightarrow n(A \cap B)=1$
$\therefore \quad P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{1}{36}$
We know that,
$
\begin{aligned}
P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\
& =\frac{12}{36}+\frac{9}{36}-\frac{1}{36}=\frac{20}{36}=\frac{5}{9}
\end{aligned}
$
$\therefore$ Probability of getting the sum multiple of 3 or 4 is $\frac{5}{9}$.
$\Rightarrow$ Probability of getting the sum neither a multiple of 3 or $4=P(\bar{A} \cap \bar{B})$
$\begin{array}{l}=P(\overline{A \cup B}) \\ =1-P(A \cup B) \\ =1-\frac{5}{9} \\ =\frac{4}{9} .\end{array}$

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Question 75 Marks

An unbiased die is thrown twice, let the event $A$ be 'odd number on first throw' and $B$ the event 'odd number on the second throw'. Check independence of the events.

Answer

Sample space of random experiment is given by
$
\begin{aligned}
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)\} \\
\therefore \quad & n(S)=36
\end{aligned}
$
Event $A =$ odd number on first throw
$\therefore \quad n(A)=18$
Event $B=$ odd number on second throw
$\therefore \quad n(B)=18$
$\begin{array}{ll}\text { Therefore, } & P(A)=\frac{n(A)}{n(S)}=\frac{18}{36}=\frac{1}{2} \\ \text { and } & P(B)=\frac{n(B)}{n(S)}=\frac{18}{36}=\frac{1}{2}\end{array}$
Also, P (odd number on both throws $)=P(A \cap B)$
$\begin{array}{l}=\frac{n(A \cap B)}{n(S)} \\ =\frac{9}{36}=\frac{1}{4}\end{array}$
Clearly, $\quad P(A \cap B)=P(A) \cdot P(B)$
Thus, events $A$ and $B$ are independent.

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