5 Marks Questions

Question 15 Marks

The probability of a person being alive in next $25$ years is $3 / 5$ and of his wife to be alive in same $25$ years is $2 / 3$. Calculate the probability of :
(1) both being alive,
(2) none being alive,
(3) at least one to be alive,
(4) only wife to be alive.

Answer

Probability of person being alive for next 25 years
$\begin{array}{l}P(A)=\frac{3}{5} \\P(\overline{A})=1-\frac{3}{5}=\frac{2}{5}\end{array}$
Probability of wife being alive for next 25 years
$\begin{array}{l}P(B)=\frac{2}{3} \\P(\overline{B})=1-\frac{2}{3}=\frac{1}{3}\end{array}$
(1) Probability of both being alive
$\begin{aligned}P=P(A \cap B) & =P(A) \times P(B) \\& =\frac{3}{5} \times \frac{2}{3}=\frac{2}{5}\end{aligned}$
(2) Probability of none being alive
$\begin{aligned}P & =P(\overline{A} \cap \overline{B}) \\& =P(\overline{A}) \times P(\overline{B}) \\& =\frac{2}{5} \times \frac{1}{3}=\frac{2}{15}\end{aligned}$
(3) Probability of at least one being alive
$\begin{aligned}P & =P(A \cap \overline{B})+P(\overline{A} \cap B)+P(A \cap B) \\& =P(A) \cdot P(\overline{B})+P(\overline{A}) \cdot P(B)+P(A) \cdot P(B) \\& =\frac{3}{5} \times \frac{1}{3}+\frac{2}{5} \times \frac{2}{3}+\frac{3}{5} \times \frac{2}{3}\end{aligned}$
$=\frac{3}{15}+\frac{4}{15}+\frac{6}{15}=\frac{13}{15}$
(4) Probability of only wife being alive
$\begin{aligned}P & =P(\overline{A} \cap B) \\& =P(\overline{A}) \times P(B) \\& =\frac{2}{5} \times \frac{2}{3}=\frac{4}{15}\end{aligned}$

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Question 25 Marks

There are three shares of $A$ in a lottery in which there are $3$ prizes and 9 blanks. In another lottery, $B$ has 2 prizes and 6 blanks. Find the ratio of probabilities of winning of $A$ and $B$.

Answer

Probability of winning of A
Total number of shares in first lottery $=3$ prizes $+$ $9$ blanks $=12$ in which A has $3$ shares.
3 shares an be chosen from 12 shares
$\begin{array}{l}={ }^{12} C_3=\frac{12 \times 11 \times 10}{3.2 .1} \\=220\end{array}$
Total number of ways of choosing of 3 shares from 9 blank shares
$\begin{array}{l}={ }^9 C_3 \\=\frac{9.8 .7}{3.2 \cdot 1}=84\end{array}$
Probability of not winning of A
$=\frac{84}{220}=\frac{21}{55}$
Probability of winning of $A$
$=1-\frac{21}{55}=\frac{35}{55}$
Probability of winning of $B$
Total number of shares in second lottery $=2$ prize +6 blanks $=8$, in which B has 2 shares.
Total number of ways of choosing 2 shares from 8 shares
$={ }^8 C_2=\frac{8.7}{2.1}=28$
Total number of ways of choosing 2 shares from 6 blank shares
$={ }^6 C_2=\frac{6.5}{2.1}=15$
Probability of not winning of $B=\frac{15}{28}$
Probability of winning of $B =1-\frac{15}{28}=\frac{13}{28}$
Ratio of winning of probabilities of A and B
$\begin{array}{l}=\frac{\frac{34}{55}}{\frac{13}{28}}=\frac{34}{55} \times \frac{28}{13}=\frac{952}{715} \\=952: 715\end{array}$

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Question 35 Marks

$A$ and $B$ played 12 games of chess in which $A$ won 6 games, two were drawn. They both decided to play three more games. Find the probability when :
(1) A wins three games,
(2) Two games were drawn,
(3) A and B wins alternately.

Answer

Probability of winning of $A=P(A)=\frac{6}{12}=\frac{1}{2}$
Probability of winning of $B=P(B)=\frac{4}{12}=\frac{1}{3}$
Probability of match drawn $=\frac{2}{12}=\frac{1}{6}$
(1) A wins three matches, its probability
$\begin{aligned}P\left(P_1 P_2 P_3\right) & =P\left(P_1\right) \times P\left(P_2\right) \times P\left(P_3\right) \\& =\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}\end{aligned}$
(2) Two games can be drawn in following ways Probability for $A = WDD,$ $DWD,$ $DDW$
$\begin{array}{l}=3\left(\frac{1}{2} \times \frac{1}{6} \times \frac{1}{6}\right) \\=\frac{3}{2} \times \frac{1}{36}=\frac{1}{24}\end{array}$
Probability for $B =$ $WDD, DWD, DDW$
$\begin{aligned}& =3\left(\frac{1}{3} \times \frac{1}{6} \times \frac{1}{6}\right)=\frac{1}{36} \\P(A+B) & =P(A)+P(B) \\& =\frac{1}{24}+\frac{1}{36} \\& =\frac{3+2}{72}=\frac{5}{72}\end{aligned}$
(3) A and B can win alternately in the following ways
$( A$ can win $) \times( B$ can win $) \times( A$ can win $)$
$=\frac{1}{2} \times \frac{1}{3} \times \frac{1}{2}=\frac{1}{12}$
Other will be in the following way
$($$B$ can win) $\times(A$ can win $) \times(B$ can win $)$
$\Rightarrow \quad \frac{1}{3} \times \frac{1}{2} \times \frac{1}{3}=\frac{1}{18}$
Hence, required probability
$=\frac{1}{12}+\frac{1}{18}=\frac{3+2}{36}=\frac{5}{36}$

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Question 45 Marks

$\text {A}$ speaks truth in $75 \%$ of the situations and $\text {B}$ in $80 \%$ of the situations. Find out in how many percent situations do they oppose each other?

Answer

$\begin{aligned}\text {Let }\quad E _1 & =\text { Event of speaking truth of } A \\ \overline{ E _1} & =\text { Event of not speaking truth of } A \\ E _2 & =\text { Event of speaking truth of } B \\ \overline{ E _2} & =\text { Event of not speaking truth of } B \end{aligned}$
$\begin{array}{l}P\left(E_1\right)=\frac{75}{100}=\frac{3}{4} \\P\left(\bar{E}_1\right)=1-P\left(E_1\right)=1-\frac{3}{4}=\frac{1}{4} \\P\left(E_2\right)=\frac{80}{100}=\frac{4}{5} \\P\left(\bar{E}_2\right)=1-P\left(E_2\right)=1-\frac{4}{5}=\frac{1}{5}\end{array}$
Let E is the event when A and B oppose each other. Now A and B can oppose each other in the following ways :
(i) A speaks truth but B does not speak truth. This event is represented by $E _1 \cap \overline{ E _2}$ or $E _1 \overline{ E _2}$.
(ii) A does not speak truth but B speaks truth. This event is represented by $\overline{ E _1} \cap E _2$ or $\overline{ E _1} E _2$.
Both events are mutually exclusive
$\begin{aligned}\Rightarrow \quad P(E) & =P\left(E_1 \overline{E}_2\right)+P\left(\overline{E}_1 E_2\right) \\& =P\left(E_1\right) \cdot P\left(\overline{E}_2\right)+P\left(\overline{E}_1\right) \cdot P\left(E_2\right) \\& =\frac{3}{4} \cdot \frac{1}{5}+\frac{1}{4} \cdot \frac{4}{5}=\frac{3}{20}+\frac{4}{20} \\& =\frac{7}{20}=\frac{35}{100}\end{aligned}$
They oppose each other in $35 \%$ of the situations.

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Question 55 Marks

$A$ and $B$ are throwing two dice one after the other. If before $B$ throwing a 7, $A$ throws 6 then $A$ wins. If $B$ throws a 7 before $A$ throws a 6 then $B$ wins. If $A$ starts throwing then prove that the probability of winning of $A$ is $\frac{30}{61}.$

Answer

Favourable outcomes for obtaining a total of 6 on two dice $=(1,5),(2,4),(3,3),(4,2)$ and $(5,1)$
Probability of getting a sum of $6=\frac{5}{36}$
Probability of not getting a sum of $6=1-\frac{5}{36}=\frac{31}{36}$
In the same way, favourable outcomes for obtaining a sum of 7 are $(1,6),(2,5),(3,4),(4,3),(5,2)$ and $(6,1)$
Probability of getting a sum of $7=\frac{6}{36}=\frac{1}{6}$
Probability of not getting a sum of $6=1-\frac{1}{6}=\frac{5}{6}$
A start then the following possibilities of winning of A are
(i) A gets a 6 in first throw whose probability is $\frac{5}{36}$
(ii) A does not get 6 in first throw, B does not get 7 in second throw and A gets 6 in third throw whose probability will be $\frac{31}{36} \times \frac{5}{6} \times \frac{5}{36}$
(iii) A does not get a 6 in first throw, B does not get 7 in second throw, A does not get 6 in third throw, B does not get a 7 in fourth throw and A gets a 6 in fifth throw whose probability will be
$\left(\frac{31}{36}\right) \times\left(\frac{5}{6}\right) \times\left(\frac{31}{36}\right) \times\left(\frac{5}{6}\right) \times \frac{5}{36}$
$=\left\{\left(\frac{31}{36}\right) \times\left(\frac{5}{6}\right)\right\}^2 \times \frac{5}{36}$
If we assume that A throws dice infinite times to win and since all these events are mutually exclusive, then by addition theorem of probability, probability of winning of $A$ is
$=\frac{5}{36}+\frac{5}{36}\left\{\left(\frac{31}{36}\right) \times\left(\frac{5}{6}\right)\right\}+\frac{5}{36}\left\{\left(\frac{31}{36}\right) \times\left(\frac{5}{6}\right)\right\}^2+\ldots$
The above series is infinite series of geometric progression. Hence sum of infinite terms
$\begin{aligned}=S_{\infty} & =\frac{a}{1-r} \\& =\frac{\frac{5}{36}}{1-\left\{\left(\frac{31}{36}\right)\times\left(\frac{5}{6}\right)\right\}}=\frac{5}{36} \times \frac{36 \times 6}{61} \\& =\frac{30}{61}\end{aligned}$

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Question 65 Marks

If two dice are thrown, then what is the probability of sum of digits obtained on them (a) greater than 8 (b) neither 7 nor 11 ?

Answer

Total outcomes on throwing of two dice is 36 .
(a) Sum of the digits greater than $8$ can be $9,10,11$ and $12$
Favourable outcomes for getting a sum of 9 are $(3,6)$, $(4,5),(5,4)$ and $(6,3)$
Probability of getting a sum of $9\left( P _1\right)=\frac{4}{36}$
Favourable outcomes for getting a sum of 10 are $(4,6),(5,5)$ and $(6,4)$
Probability of getting a sum of $10\left( P _2\right)=\frac{3}{36}$
Favourable outcomes for getting a sum of 11 are $(5,6)$ and $(6,5)$
Probability of getting a sum of $11\left( P _3\right)=\frac{2}{36}$
Favourable outcomes for getting a sum of 12 is $(6,6)$
Probability of getting a sum of $12\left( P _4\right)=\frac{1}{36}$
From addition theorem of probability as these events are exclusive
$\begin{aligned}& =P_1+P_2+P_3+P_4 \\& =\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36} \\\frac{10}{36} & =\frac{5}{18}\end{aligned}$
(b) Favourable outcomes for obtaing a sum of 7 are $(1,6),(2,5),(3,4),(4,3),(5,2)$ and $(6,1)$
Probability of getting a sum of $7\left( P _1\right)=\frac{6}{36}$
Favourable outcomes for obtaining a sum of 11 will be $(5,6)$ and $(6,5)$
Probability of getting a sum of $11\left( P _2\right)=\frac{2}{36}$
Probability of getting a sum of 7 or $11=\frac{6}{36}+\frac{2}{36}$
$=\frac{8}{36}=\frac{2}{9}$
Probability of not getting a sum of 7 or 11
$=1-\frac{2}{9}=\frac{9-2}{9}=\frac{7}{9}$

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Question 75 Marks

There are three independent witnesses of any event that has occurred. A speaks truth three times out of four times. B speaks truth four times out of five times and C speaks truth five times out of six times. What is the probability of all the witnesses speaking in favour of the event with majority?

Answer

Let $E_1, E_2$ and $E_3$ represents events of speaking truth by $A , B$ and $C$ respectively.
$\begin{array}{l}P\left(E_1\right)=\frac{3}{4}, P\left(E_2\right)=\frac{4}{5}, P\left(E_3\right)=\frac{5}{6} \\\therefore \quad P\left(\overline{E}_1\right)=1-\frac{3}{4}=\frac{1}{4} \\P\left(\overline{E}_2\right)=1-\frac{4}{5}=\frac{1}{5} \\P\left(\overline{E}_3\right)=1-\frac{5}{6}=\frac{1}{6}\end{array}$
Out of $A , B$ and $C,$ the following conditions are there in which at least two are speaking truth.
$(\text i)$ $E _1 E _2 E _3$
$(\text {ii}$) $E_1 E_2 \bar{E}_3$
$(\text {iii}$) $E _1 \overline{ E }_2 E _3$
$(\text {iv}$) $\overline{ E }_1 E _2 E _3$
$P (i)= P \left( E _1 E _2 E _3\right)= P \left( E _1\right) \cdot P \left( E _2\right) \cdot P \left( E _3\right)$
$\begin{aligned}& =\frac{3}{4} \times \frac{4}{5} \times \frac{5}{6}=\frac{1}{2} \\P(ii) & =P\left(E_1 E_2 \overline{E}_3\right)=P\left(E_1\right) \cdot P\left(E_2\right) \cdot P\left(\overline{E}_3\right) \\& =\frac{3}{4} \times \frac{4}{5} \times \frac{1}{6}=\frac{1}{10} \\P(iii) & =P\left(E_1 \overline{E}_2E_3\right)=P\left(E_1\right) \cdot P\left(\overline{E}_2\right) \cdot P\left(E_3\right) \\& =\frac{3}{4} \times \frac{1}{5} \times \frac{5}{6}=\frac{1}{8} \\P(iv) & =P\left(\overline{E}_1 E_2E_3\right)=P\left(\overline{E}_1\right) \cdot P\left(E_2\right) \cdot P\left(E_3\right) \\& =\frac{1}{4} \times \frac{4}{5} \times \frac{5}{6}=\frac{1}{6}\end{aligned}$
Since all the four events are mutually exclusive, therefore by addition theorem of probability, required probability
$=\frac{1}{2}+\frac{1}{10}+\frac{1}{8}+\frac{1}{6}=\frac{107}{120}$

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