Maths Solve the Following Question.(4 Marks)

∴ Sample space S = {BB, BG, GB, GG}

(i) A: First child is a girl.

∴ A = {GB, GG}

$\therefore P(A)=\frac{2}{4}=\frac{1}{2}$

B: Second child is a girl.

∴ B = {BG, GG}

∴ A ∩ B = {GG}

$\therefore P(A \cap B)=\frac{1}{4}$

Required probability

$P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$

ii. A: At least one of the children is a girl.

∴ A = {GG, GB, BG}

$\therefore P(A)=\frac{3}{4}$

B: both children are girls. B = {GG}

$\therefore P(B)=\frac{1}{4}$

Also, A ∩ B = B

$P ( B / A )=\frac{ P ( B \cap A )}{ P ( A )}=\frac{ P ( B )}{ P ( A )}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

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)}

∴ n(S) = 36

Let event A: The sum is 5 in a trial.

A = {(2, 3), (3, 2), (1, 4), (4, 1)}

$\therefore P(A)=\frac{4}{36}=\frac{1}{9}$

Let event B: The sum is 7 in a trial.

B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}

$\therefore P(B)=\frac{6}{36}=\frac{1}{6}$

Let event C: Neither sum is 5 nor 7. P(C) = 1 – P(A) – P(B)

$\begin{aligned} & =1-\frac{1}{9}-\frac{1}{6} \\ & =\frac{26}{36}\end{aligned}$

Let the sum of 5 appear in the nth trial for the first time and the sum of 7 has not occurred in the first (n – 1) trials.

Probability of this event $=[ P ( C )]^{ n -1} P ( A )$

Required probability $=\sum_{n=1}^{\infty}\left(\frac{26}{36}\right)^{n-1}\left(\frac{1}{9}\right)$

$=\frac{1}{9}\left(\frac{1}{1-\frac{26}{36}}\right)$

$=\frac{1}{9}\left(\frac{18}{5}\right)=\frac{2}{5}$

event B: The student is a boy,

event G: The student is a girl.

Since the ratio of boys to girls is 3 : 2 and 3 girls out of 500 and 2 boys out of 50 are good singers.

$\begin{aligned} & P ( B )=\frac{3}{5}, P ( G )=\frac{2}{5}, P ( S / G )=\frac{3}{500}, \\ & P ( S / B )=\frac{2}{50} . \\ & P ( S )= P ( G ) \times P ( S / G )+ P ( B ) \times P ( S / B )\end{aligned}$

$\begin{aligned} & =\frac{2}{5} \times \frac{3}{500}+\frac{3}{5} \times \frac{2}{50} \\ & =\frac{2 \times 3}{5}\left(\frac{1}{500}+\frac{1}{50}\right) \\ & =\frac{6}{5} \times \frac{11}{500}\end{aligned}$

$=\frac{33}{1250}$

Required probability $= P ( G / S )$

By Bayes' theorem,

$\begin{aligned} P(G / S) & =\frac{P(G) P(S / G)}{P(S)} \\ & =\frac{\frac{2}{5} \times \frac{3}{500}}{\frac{33}{1250}}=\frac{1}{11}\end{aligned}$

event Q: Q become principal,

event R: R become principal,

event E: Subject IT is introduced.

$\begin{aligned} & \text { Given, } P(P)=\frac{2}{5} \\ & P(Q)=\frac{2}{5} \\ & P(R)=\frac{1}{5} \\ & P(E / P)=\frac{1}{2} \\ & P(E / Q)=\frac{1}{3} \\ & P(E / R)=\frac{1}{4}\end{aligned}$

(a) Required probability

P(E) = P(P) P(E/P) + P(Q) P(E/Q) + P(R) P(E/R)

$\begin{aligned} & =\frac{2}{5} \times \frac{1}{2}+\frac{2}{5} \times \frac{1}{3}+\frac{1}{5} \times \frac{1}{4} \\ & =\frac{1}{5}+\frac{2}{15}+\frac{1}{20} \\ & =\frac{12+8+3}{60} \\ & =\frac{23}{60}\end{aligned}$

2. Required probability = P(Q/E) By Bayes’ theorem,

$\begin{aligned} & P ( Q / E )=\frac{P(Q) P(E / Q)}{P(E)} \\ & =\frac{\frac{2}{5} \times \frac{1}{3}}{\frac{23}{60}} \\ & =\frac{8}{23}\end{aligned}$

∴ n(S) = 10

A: Number is more than 3.

A = {4, 5, 6, 7, 8, 9, 10}

∴ n(A) = 7

$\therefore P ( A )=\frac{n(A)}{n(S)}=\frac{7}{10}$

B: Number is even. B = {2, 4, 6, 8, 10}

∴ A ∩ B = {4, 6, 8, 10}

∴ n(A ∩ B) = 4

$\therefore P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{4}{10}$

Required probability = P(B/A)

$\begin{aligned} & =\frac{P(A \cap B)}{P(A)} \\ & =\frac{\left(\frac{4}{10}\right)}{\left(\frac{7}{10}\right)} \\ & =\frac{4}{7}\end{aligned}$

event B: Four negative numbers are chosen,

event C: Two positive and two negative numbers are chosen.

Since four numbers are chosen without replacement,

n(A) = 6 × 5 × 4 × 3 = 360

n(B) = 8 × 7 × 6 × 5 = 1680

In event C, four numbers are to be chosen without replacement such that two numbers are positive and two numbers ate negative. This can be done in following ways:

+ + – – OR + – + – OR + – – + OR – + – + OR – – + + OR – + + –

∴ n(C) = 6 × 5 × 8 × 7 + 6 × 8 × 5 × 7 + 6 × 8 × 7 × 5 + 8 × 6 × 7 × 5 + 6 × 5 × 8 × 7 + 8 × 6 × 5 × 7

= 6 × (8 × 7 × 6 × 5)

=10080

Here, total number of numbers = 14

∴ n(S) = 14 × 13 × 12 × 11 = 24024

Since A, B, C are mutually exclusive events,

Required probability = P(A) + P(B) + P(C)

$\begin{aligned} & =\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}+\frac{n(C)}{n(S)} \\ & =\frac{360+1680+10080}{24024} \\ & =\frac{505}{1001}\end{aligned}$

Let $P\left(H^{\prime}\right)=P($ husband dies before he is 85$)=\frac{7}{7+5}=\frac{7}{12}$

So, the probability that the husband would be alive till age 85

$P(H)=1-P\left(H^{\prime}\right)=1-\frac{7}{12}=\frac{5}{12}$

Similarly, P(W’) = P(Wife dies before she is 81) Since the odds against wife will be alive till she is 81 are 5 : 3.

$\therefore P\left(W^{\prime}\right)=\frac{5}{5+3}=\frac{5}{8}$

So, the probability that the wife would be alive till age 81

$P(W)=1-P\left(W^{\prime}\right)=1-\frac{5}{8}=\frac{3}{8}$

(i) Required probability

P(H ∪ W) = P(H) + P(W) – P(H ∩ W)

Since H and W are independent events,

P(H ∩ W) = P(H) . P(W)

∴ Required probability = P(H) + P(W) – P(H) . P(W)

$\begin{aligned} & =\frac{5}{12}+\frac{3}{8}-\frac{5}{12} \times \frac{3}{8} \\ & =\frac{40+36-15}{96} \\ & =\frac{61}{96}\end{aligned}$

2. Required probability = P(H ∩ W’) + P(H’ ∩ W) Since H and W are independent events, H’ and W’ are also independent events.

∴ Required probability = P(H) . P(W’) + P(H’) . P(W)

$\begin{aligned} & =\frac{5}{12} \times \frac{5}{8}+\frac{7}{12} \times \frac{3}{8} \\ & =\frac{25+21}{96} \\ & =\frac{46}{96} \\ & =\frac{23}{48}\end{aligned}$

The odds against X solving a problem are 8 : 6.

Let $P\left(X^{\prime}\right)=P(X$ does not solve the problem $)=\frac{8}{8+6}=\frac{8}{14}$

So, the probability that X solves the problem

$P(X)=1-P\left(X^{\prime}\right)=1-\frac{8}{14}=\frac{6}{14}$

Similarly, let P(Y) = P(Y solves the problem)

Since odds in favour of Y solving the problem are 14 : 16,

$P(Y)=\frac{14}{14+16}=\frac{14}{30}$

So, the probability that Y does not solve the problem P(Y’) = 1 – P(Y)

$\begin{aligned} & =1-\frac{14}{30} \\ & =\frac{16}{30}\end{aligned}$

(i) Required probability

P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)

Since X and Y are independent events,

P(X ∩ Y) = P(X) . P(Y)

∴ Required probability = P(X) + P(Y) – P(X) . P(Y)

$\begin{aligned} & =\frac{6}{14}+\frac{14}{30}-\frac{6}{14} \times \frac{14}{30} \\ & =\frac{73}{105}\end{aligned}$

2.Required probability = P(X’ ∩ Y’)

Since X and Y are independent events, X’ and Y’ are also independent events.

∴ Required probability = P(X’) . P(Y’)

$\begin{aligned} & =\frac{8}{14} \times \frac{16}{30} \\ & =\frac{32}{105}\end{aligned}$

event B: Message sent on sports by group II,

event C: Message sent on sports by group III,

event E: Message is opened.

Given that the probabilities that Group I, Group II and Group III sending the messages on

sports are $\frac{2}{5}, \frac{1}{2}$ and $\frac{2}{3}$ respectively and the probability of opening the messages by Group

I, Group II and Group III are $\frac{1}{2}, \frac{1}{4}$ and $\frac{1}{4}$ respectively.

$\begin{aligned} & \therefore P(A)=\frac{2}{5} \\ & P(B)=\frac{1}{2} \\ & P(C)=\frac{2}{3} \\ & P(E / A)=\frac{1}{2} \\ & P(E / B)=\frac{1}{4} \\ & P(E / C)=\frac{1}{4}\end{aligned}$

Required probability = P(C/E) By Baye’s theorem

$\begin{aligned} & P ( C / E ) \\ & =\frac{ P ( C ) P ( E / C )}{ P ( A ) P ( E / A )+ P ( B ) P ( E / B )+ P ( C ) P ( E / C )} \\ & =\frac{\frac{2}{3} \times \frac{1}{4}}{\frac{2}{5} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{4}+\frac{2}{3} \times \frac{1}{4}}\end{aligned}$

$\begin{aligned} & =\frac{\frac{1}{6} \text {}}{\frac{1}{5}+\frac{1}{8}+\frac{1}{6}} \\ & =\frac{\frac{1}{6}}{\frac{59}{120}}=\frac{20}{59}\end{aligned}$

event B: Fake coin is tossed

and event H: Head occur.

Clearly, a fair coin has one head.

$\therefore$ Probability that head occur under the condition that the fair coin is tossed $=P(H / A)=\frac{1}{2}$

Fake coin has two heads.

∴ Probability that head occur under the condition that the fake coin is tossed = P(H/B) = 1 n(A) = 2, n(B) = 1, n(S) = 3

$\begin{aligned} & \therefore P ( A )=\frac{n(A)}{n(S)}=\frac{2}{3} \\ & P ( B )=\frac{n(B)}{n(S)}=\frac{1}{3}\end{aligned}$

(i) Required probability P(H) = P(A) P(H/A) + P(B) P(H/B)

$\begin{aligned} & =\frac{2}{3} \times \frac{1}{2}+\frac{1}{3} \times 1 \\ & =\frac{1}{3}+\frac{1}{3} \\ & =\frac{2}{3}\end{aligned}$

2. Required probability = P(B/H) By Baye’s theorem

$\begin{aligned} P(B / H) & =\frac{P(B) P(H / B)}{P(H)} \\ & =\frac{\frac{1}{3} \times 1}{\frac{2}{3}}\end{aligned}$

$=\frac{1}{2}$

$P(S)=\frac{0.5}{100}=0.005$

∴ P(S’) = 1 – P(S) = 1 – 0.005 = 0.995 Since a probability of getting a positive result when applied to a person suffering from a disease is 0.95 and probability of getting positive result when applied to a non sufferer is 0.10.

∴ P(T/S) = 0.95 and P(T/S’) = 0.10

∴ P(T) = P(S) P(T/S) + P(S’) P(T/S’) = 0.005 × 0.95 + 0.995 × 0.10 = 0.10425

∴ P(T’) = 1 – P(T) = 1 – 0.10425 = 0.8958

(i) Required probability = P(S/T)

By Bayes’ theorem,

$P ( S / T )=\frac{ P ( S ) P ( T / S )}{ P ( T ) }$

$=\frac{0.005 \times 0.95}{0.10425}=\frac{0.00475}{0.10425}$

2. P(T’/S’) = 1 – 0.1 = 0.9

Required probability = P(S’/T’)

By Bayes’ theorem

$\begin{aligned} P \left( S ^{\prime} / T ^{\prime}\right) & =\frac{ P \left( S ^{\prime}\right) P \left( T ^{\prime} / S ^{\prime}\right)}{ P \left( T ^{\prime}\right)} \\ & =\frac{0.995 \times 0.9}{0.8958} \\ & =\frac{0.8955}{0.8958}\end{aligned}$

∴ n(S) = 100

Let event N: Women are from neighbouring town, event W: Women are from same town and event G: Women are graduates. Number of women from neighbouring town,

n(N) = 75

Number of women from same town,

n(W) = 25

$\begin{aligned} & \therefore P ( N )=\frac{n(N)}{n(S)}=\frac{75}{100} \text { and } \\ & P ( W )=\frac{n(W)}{n(S)}=\frac{25}{100}\end{aligned}$

P(G/N), P(G/W) represent probabilities that woman is graduate given that she is from neighbouring town or same town respectively.

$\begin{aligned} & \therefore P(G / N)=\frac{n(G / N)}{n(S)}=\frac{83}{100} \text { and } \\ & P(G / W)=\frac{n(G / W)}{n(S)}=\frac{48}{100}\end{aligned}$

By Bayes’ theorem, the probability that a women selected at random is a graduate from the same town, is given by

$P ( W / G )=\frac{ P ( W ) P ( G / W )}{ P ( W ) \cdot P ( G / W )+ P ( N ) \cdot P ( G / N )}$

$=\frac{\left(\frac{25}{100}\right) \cdot\left(\frac{48}{100}\right)}{\left(\frac{25}{100}\right) \cdot\left(\frac{48}{100}\right)+\left(\frac{75}{100}\right) \cdot\left(\frac{83}{100}\right)}$

$\begin{aligned} & =\frac{25 \times 48}{(25 \times 48)+(75 \times 83)} \\ & =\frac{48}{48+249}\end{aligned}$

$=\frac{16}{99}$

event $A _2:$ The ball is drawn from the 2 nd box.

$\therefore P\left(A_1\right)=\frac{1}{2}, P\left(A_2\right)=\frac{1}{2}$

Let event B: The ball drawn is pink. There are 5 balls in the 1st box, of which 3 are pin

$\therefore P\left(B / A_1\right)=\frac{3}{5}$

There are 9 balls in the 2nd box, of which 5 are pink.

$\therefore P \left( B / A _2\right)=\frac{5}{9}$

(i) By Bayes’ theorem,

the probability that a pink ball is drawn from the first box, is given by

$P\left(A_1 / B\right)=\frac{P\left(A_1\right) \cdot P\left(B / A_1\right)}{P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right) \cdot P\left(B / A_2\right)}$

$=\frac{\frac{1}{2} \times \frac{3}{5}}{\frac{1}{2} \times \frac{3}{5}+\frac{1}{2} \times \frac{5}{9}}$

$=\frac{\frac{3}{10}}{\frac{3}{10}+\frac{5}{18}}=\frac{\frac{3}{10}}{\frac{27+25}{90}}=\frac{\frac{3}{10}}{\frac{52}{90}}=\frac{27}{52}$

2. By Bayes’ theorem,

the probability that a pink ball is drawn from the second box, is given by

$\begin{aligned} & P\left(A_2 / B\right) \\ & =\frac{P\left(A_2\right) \cdot P\left(B / A_2\right)}{P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right) \cdot P\left(B / A_2\right)}\end{aligned}$

$=\frac{\frac{1}{2} \times \frac{5}{9}}{\frac{1}{2} \times \frac{3}{5}+\frac{1}{2} \times \frac{5}{9}}$

$=\frac{\frac{5}{18}}{\frac{3}{10}+\frac{5}{18}}=\frac{\frac{5}{18}}{\frac{27+25}{90}}=\frac{\frac{5}{18}}{\frac{52}{90}}=\frac{25}{52}$

(a) Let event A: First ball drawn is red.

$\therefore P(A)=\frac{{ }^{10} C _1}{{ }^{25} C _1}=\frac{10}{25}=\frac{2}{5}$

Let event B: Second ball drawn is green. Since the first red ball is not replaced in the box, we now have 24 balls, out of which 15 are green.

∴ Probability that the second ball is green under the condition that the first red ball is not

replaced in the box $=P(B / A)=\frac{{ }^{15} C _1}{{ }^{24} C _1}=\frac{15}{24}=\frac{5}{8}$

∴ Required probability = P(A ∩ B) = P(B/A) . P(A)

$\begin{aligned} & =\frac{2}{5} \times \frac{5}{8} \\ & =\frac{1}{4}\end{aligned}$

2. To find the probability that one ball is red and the other is green, there are two possibilities:

First ball is red and second ball is green.

OR

The first ball is the green and the second ball is red.

From above, we get

$P$ (First ball is red and second ball is green) $=\frac{1}{4}$

Similarly,

$P ($ First ball is green and second ball is red $)=\frac{{ }^{15} C _1}{{ }^{25} C _1} \times \frac{{ }^{10} C _1}{{ }^{24} C _1}=\frac{15}{25} \times \frac{10}{24}=\frac{1}{4}$

∴ Required probability = P(First ball is red and second ball is green) + P(First ball is green and second ball is red)

$\begin{aligned} & =\frac{1}{4}+\frac{1}{4} \\ & =\frac{1}{2}\end{aligned}$

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)}

∴ n(S) = 36 Let event A: Getting 6 on the first die.

∴ A = {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ n(A) = 6

$\therefore P ( A )=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

Let event B : Gettting 2 on the second die.

∴ B = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)}

∴ n(B) = 6

$\therefore P ( B )=\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

Now, A ∩ B = {(6, 2)}

∴ n(A ∩ B) = 1

$\begin{aligned} & \therefore P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{1}{36} \ldots \ldots \text { (i) } \\ & P(A) \times P(B)=\frac{1}{6} \times \frac{1}{6}=\frac{1}{36} \ldots \ldots .(\text {ii) }\end{aligned}$

From (i) and (ii), we get

P(A ∩ B) = P(A) × P(B)

∴ A and B are independent events.

event B: B can hit the target,

event C: C can hit the target.

$\therefore \quad P ( A )=\frac{3}{4}, P ( B )=\frac{1}{2}, P ( C )=\frac{5}{8}$

$\therefore \quad P \left( A ^{\prime}\right)=1- P ( A )=1-\frac{3}{4}=\frac{1}{4}$

$P\left(B^{\prime}\right)=1-P(B)=1-\frac{1}{2}=\frac{1}{2}$

$P\left(C^{\prime}\right)=1-P(C)=1-\frac{5}{8}=\frac{3}{8}$

Since A, B, C are independent events,

A’, B’, C’ are also independent events.

(a) Let event W: Target is hit exactly by one of them.

$\begin{array}{r} P ( W )= P \left( A \cap B ^{\prime} \cap C ^{\prime}\right) \cup P \left( A ^{\prime} \cap B \cap C ^{\prime}\right) \\ \cup P \left( A ^{\prime} \cap B ^{\prime} \cap C \right) \\ = P ( A ) \cdot P ^{\prime}\left( B ^{\prime}\right) \cdot P \left( C ^{\prime}\right)+ P \left( A ^{\prime}\right) \cdot P ( B ) \cdot P \left( C ^{\prime}\right)\end{array}$

$+ P \left( A ^{\prime}\right) \cdot P \left( B ^{\prime}\right) \cdot P ( C )$

$\begin{aligned} & =\left(\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}\right)+\left(\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8}\right)+\left(\frac{1}{4} \times \frac{1}{2} \times \frac{5}{8}\right) \\ & =\frac{9}{64}+\frac{3}{64}+\frac{5}{64}=\frac{17}{64}\end{aligned}$

Let event X: Target is not hit by any one of them.

∴ P(X) = P(A’ ∩ B’ ∩ C’)

= P(A’) P(B’) P(C’)

$\begin{aligned} & =\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8} \\ & =\frac{3}{64}\end{aligned}$

Let event Y: Target is hit.

∴ P(Y) = 1 – P(target is not hit by any one of them)

$\begin{aligned} & =1-\frac{3}{64} \\ & =\frac{61}{64}\end{aligned}$

Let event Z: Target is hit by exactly two of them.

$\begin{aligned} P(Z)=P\left(A \cap B \cap C^{\prime}\right) \cup P\left(A \cap B^{\prime} \cap C\right) \\ \cup P\left(A^{\prime} \cap B \cap C\right) \\ =P(A) \cdot P(B) \cdot P\left(C^{\prime}\right)+P(A) \cdot P\left(B^{\prime}\right) \cdot P(C)\end{aligned}$

$+ P \left( A ^{\prime}\right) \cdot P ( B ) \cdot P ( C )$

$\begin{aligned} & =\left(\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}\right)+\left(\frac{3}{4} \times \frac{1}{2} \times \frac{5}{8}\right)+\left(\frac{1}{4} \times \frac{1}{2} \times \frac{5}{8}\right) \\ & =\frac{9}{64}+\frac{15}{64}+\frac{5}{64}=\frac{29}{64}\end{aligned}$

∴ n(S) = 50

Let event A: The number drawn is divisible by 2.

∴ A = {2, 4, 6, 8, 10, …, 48, 50}

∴ n(A) = 25

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{25}{50}$

Let event B: The number drawn is divisible by 3.

B = {3, 6, 9, 12, …, 48}

∴ n(B) = 16

$\therefore P(B)=\frac{n(B)}{n(S)}=\frac{16}{50}$

Let event C: The number drawn is divisible by 10.

C = {10, 20, 30, 40, 50}

∴ n(C) = 5

$\therefore P (C)=\frac{n(C)}{n(S)}=\frac{5}{50}$

Now, A ∩ B = {6, 12, 18, 24, 30, 36, 42, 48}

∴ n(A ∩ B) = 8

$\therefore P ( A \cap B )=\frac{n(A \cap B)}{n(S)}=\frac{8}{50}$

B ∩ C = {30}

∴ n(B ∩ C) = 1

$\therefore P ( B \cap C )=\frac{ n ( B \cap C )}{ n ( S )}=\frac{1}{50}$

A ∩ C = {10, 20, 30, 40, 50} ∴ n(A ∩ C) = 5

$\therefore P(A \cap C)=\frac{n(A \cap C)}{n(S)}=\frac{5}{50}$

A ∩ B ∩ C = {30}

∴ n(A ∩ B ∩ C) = 1

$\therefore P ( A \cap B \cap C )=\frac{n(A \cap B \cap C)}{n(S)}=\frac{1}{50}$

∴ P(the number is divisible by 2 or 3 or 10) P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

$\begin{aligned} & =\frac{25}{50}+\frac{16}{50}+\frac{5}{50}-\frac{8}{50}-\frac{1}{50}-\frac{5}{50}+\frac{1}{50} \\ & =\frac{33}{50}\end{aligned}$

$\therefore 4$ persons can be selected from 11 persons in ${ }^{11} C _4$ ways.

$\therefore n ( S )={ }^{11} C _4$

(i) Let event A: No child is selected.

$\therefore 4$ persons can be selected from 4 men and 4 women, i.e., from 8 persons in ${ }^8 C _4$ ways.

$\begin{aligned} \therefore n ( A ) & ={ }^8 C _4 \\ P ( A ) & =\frac{ n ( A )}{ n ( S )} \\ & =\frac{{ }^8 C _4}{{ }^{11} C _4} \\ & =\frac{8 \times 7 \times 6 \times 5}{11 \times 10 \times 9 \times 8} \\ & =\frac{1680}{7920} \\ & =\frac{7}{33}\end{aligned}$

2. Let event B: Exactly 2 men are selected.

$\therefore 2$ men are selected from 4 men in ${ }^4 C _2$ ways, and remaining 2 persons are selected from

7 persons (i.e., 4 women and 3 children) in ${ }^7 C _2$ ways.

$\begin{aligned} n ( A ) & ={ }^4 C _2 \times{ }^7 C _2 \\ P ( A ) & =\frac{ n ( A )}{ n ( S )} \\ & =\frac{{ }^4 C _2 \times{ }^7 C _2}{{ }^{11} C _4} \\ & =\frac{\frac{4 \times 3}{2} \times \frac{7 \times 6}{2}}{\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}} \\ & =\frac{126}{330} \\ & =\frac{21}{55}\end{aligned}$

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4,4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}

∴ n(S) = 30

Let event A: The sum of the numbers on the upper faces of the dice is divisible by 2.

A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4)}

∴ n(A) = 15

$\therefore P ( A )=\frac{n(A)}{n(S)}=\frac{15}{30}$

Let event B: Sum of the numbers on the upper faces of the dice is divisible by 3.

B = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3)}

∴ n(B) = 10

$\therefore P(B)=\frac{n(B)}{n(S)}=\frac{10}{30}$

Now,

A ∩ B = {(1, 5), (2,4), (3, 3), (4, 2), (5, 1)}

∴ n(A ∩ B) = 5

$\therefore P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{5}{30}$

∴ Required probability P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

$\begin{aligned} & =\frac{15}{30}+\frac{10}{30}-\frac{5}{30} \\ & =\frac{20}{30} \\ & =\frac{2}{3}\end{aligned}$

∴ n(S) = 50

Let event A: The number drawn is divisible by 2.

∴ A = {2, 4, 6, 8, 10, …, 48, 50}

∴ n(A) = 25

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{25}{50}$

Let event B: The number drawn is divisible by 3.

B = {3, 6, 9, 12, …, 48}

∴ n(B) = 16

$\therefore P(B)=\frac{n(B)}{n(S)}=\frac{16}{50}$

Let event C: The number drawn is divisible by 10.

C = {10, 20, 30, 40, 50}

∴ n(C) = 5

$\therefore P (C)=\frac{n(C)}{n(S)}=\frac{5}{50}$

Now, A ∩ B = {6, 12, 18, 24, 30, 36, 42, 48}

∴ n(A ∩ B) = 8

$\therefore P ( A \cap B )=\frac{n(A \cap B)}{n(S)}=\frac{8}{50}$

B ∩ C = {30}

∴ n(B ∩ C) = 1

$\therefore P ( B \cap C )=\frac{ n ( B \cap C )}{ n ( S )}=\frac{1}{50}$

A ∩ C = {10, 20, 30, 40, 50} ∴ n(A ∩ C) = 5

$\therefore P(A \cap C)=\frac{n(A \cap C)}{n(S)}=\frac{5}{50}$

A ∩ B ∩ C = {30}

∴ n(A ∩ B ∩ C) = 1

$\therefore P ( A \cap B \cap C )=\frac{n(A \cap B \cap C)}{n(S)}=\frac{1}{50}$

∴ P(the number is divisible by 2 or 3 or 10) P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

$\begin{aligned} & =\frac{25}{50}+\frac{16}{50}+\frac{5}{50}-\frac{8}{50}-\frac{1}{50}-\frac{5}{50}+\frac{1}{50} \\ & =\frac{33}{50}\end{aligned}$

$\therefore 4$ persons can be selected from 11 persons in ${ }^{11} C _4$ ways.

$\therefore n ( S )={ }^{11} C _4$

(i) Let event A: No child is selected.

$\therefore 4$ persons can be selected from 4 men and 4 women, i.e., from 8 persons in ${ }^8 C _4$ ways.

$\begin{aligned} \therefore n ( A ) & ={ }^8 C _4 \\ P ( A ) & =\frac{ n ( A )}{ n ( S )} \\ & =\frac{{ }^8 C _4}{{ }^{11} C _4} \\ & =\frac{8 \times 7 \times 6 \times 5}{11 \times 10 \times 9 \times 8} \\ & =\frac{1680}{7920} \\ & =\frac{7}{33}\end{aligned}$

2. Let event B: Exactly 2 men are selected.

$\therefore 2$ men are selected from 4 men in ${ }^4 C _2$ ways, and remaining 2 persons are selected from

7 persons (i.e., 4 women and 3 children) in ${ }^7 C _2$ ways.

$\begin{aligned} n ( A ) & ={ }^4 C _2 \times{ }^7 C _2 \\ P ( A ) & =\frac{ n ( A )}{ n ( S )} \\ & =\frac{{ }^4 C _2 \times{ }^7 C _2}{{ }^{11} C _4} \\ & =\frac{\frac{4 \times 3}{2} \times \frac{7 \times 6}{2}}{\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}} \\ & =\frac{126}{330} \\ & =\frac{21}{55}\end{aligned}$

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4,4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}

∴ n(S) = 30

Let event A: The sum of the numbers on the upper faces of the dice is divisible by 2.

A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4)}

∴ n(A) = 15

$\therefore P ( A )=\frac{n(A)}{n(S)}=\frac{15}{30}$

Let event B: Sum of the numbers on the upper faces of the dice is divisible by 3.

B = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3)}

∴ n(B) = 10

$\therefore P(B)=\frac{n(B)}{n(S)}=\frac{10}{30}$

Now,

A ∩ B = {(1, 5), (2,4), (3, 3), (4, 2), (5, 1)}

∴ n(A ∩ B) = 5

$\therefore P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{5}{30}$

∴ Required probability P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

$\begin{aligned} & =\frac{15}{30}+\frac{10}{30}-\frac{5}{30} \\ & =\frac{20}{30} \\ & =\frac{2}{3}\end{aligned}$

These letters can be arranged among themselves in ${ }^6 P _6=6$ ! ways.

∴ n(S) = 6!

(a) Let event A: Vowels are always together.

The word MOTHER consists of 2 vowels (O, E) and 4 consonants (M, T, H, R).

2 vowels can be arranged among themselves in ${ }^2 P _2=2$ ! ways.

Let us consider 2 vowels as one group.

This one group with 4 consonants can be arranged in ${ }^5 P _5=5$ ! ways.

$\begin{aligned} & \therefore n ( A )=2 ! \times 5 ! \\ & \therefore P ( A )=\frac{n(A)}{n(S)}=\frac{2 ! \times 5 !}{6 !}=\frac{1}{3}\end{aligned}$

Let event B: Vowels are never together.

4 consonants create 5 gaps, in which vowels are arranged.

Consider the following arrangement of consonants

_C_C_C_C_

2 vowels can be arranged in 5 gaps in ${ }^5 P _2$ ways.

Also 4 consonants can be arranged among themselves in ${ }^4 P _4=4$ ! ways.

$\begin{aligned} & \therefore n ( B )=4 ! \times{ }^5 P _2 \\ & \therefore P ( B )=\frac{n(B)}{n(S)}=\frac{4 ! \times^5 P_2}{6 !}=\frac{4 ! \times 5 \times 4}{6 \times 5 \times 4 !}=\frac{2}{3}\end{aligned}$

Let event C: Word begin with O and end with T. Thus first and last letters can be arranged in one way each and the remaining 4 letters can

be arranged in ${ }^4 P _4=4$ ! ways

$\begin{aligned} & \therefore n ( C )=4 ! \times 1 \times 1=4 ! \\ & \therefore P ( C )=\frac{n(C)}{n(S)}=\frac{4 !}{6 !}=\frac{1}{30}\end{aligned}$

Let event D: Word starts with a vowel and ends with a consonant.

There are 2 vowels and 4 consonants in the word MOTHER.

∴ The first place can be arranged in 2 different ways and the last place can be arranged in 4 different ways.

Now, the remaining 4 letters ( 3 consonants and 1 vowel) can be arranged in ${ }^4 P_4=4$ !

ways.

$\begin{aligned} & \therefore n ( D )=2 \times 4 \times 4 ! \\ & \therefore P ( D )=\frac{n(D)}{n(S)}=\frac{2 \times 4 \times 4 !}{6 !}=\frac{4}{15}\end{aligned}$

2 balls can be drawn from 9 balls with replacement in ${ }^9 C _1 \times{ }^9 C _1$ ways.

$\therefore n ( S )={ }^9 C _1 \times{ }^9 C _1=9 \times 9=81$

Let event A: Balls drawn are red.

2 red balls can be drawn from 5 red balls with replacement in ${ }^5 C _1 \times{ }^5 C _1$ ways.

$\begin{aligned} & \therefore n ( A )={ }^5 C _1 \times{ }^5 C _1=5 \times 5=25 \\ & \therefore P(A)=\frac{n(A)}{n(S)}=\frac{25}{81}\end{aligned}$

2 balls can be drawn from 9 balls without replacement in ${ }^9 C_1 \times{ }^8 C_1$ ways.

$\therefore n ( S )={ }^9 C_1 \times{ }^8 C_1=9 \times 8=72$

2 red balls can be drawn from 5 red balls without replacement in ${ }^5 C _1 \times{ }^4 C _1$ ways.

$\begin{aligned} & \therefore n ( B )={ }^5 C _1 \times{ }^4 C _1=5 \times 4=20 \\ & \therefore P ( B )=\frac{n(B)}{n(S)}=\frac{20}{72}=\frac{5}{18}\end{aligned}$

Let event B: Number on the ticket is a perfect square.

∴ B = (1, 4, 9, 16, 25, 36, 49, 64}

∴ n(B) = 8

$\therefore P ( B )=\frac{n(B)}{n(S)}=\frac{8}{75}$

Let event C: Number on the ticket is a prime number.

∴C = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}

Event D : Self

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)}

∴ n(S) = 36

(a) Let event A: Sum of the numbers on uppermost face is 5.

∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)}

∴ n(A) = 4

$\therefore P ( A )=\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}$

Let event B: Sum of the numbers on uppermost face is at least 8 (i.e., 8 or more than 8)

∴ B = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ n(B) = 15

$\therefore P ( B )=\frac{n(B)}{n(S)}=\frac{15}{36}=\frac{5}{12}$

Let event C: First throw gives a multiple of 2 and second throw gives a multiple of 3.

∴ C = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6)}

∴ n(C) = 6

$\therefore P(C)=\frac{n(C)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

Let event D: The product of the numbers on uppermost face is 12.

∴ D = {(2, 6), (3, 4), (4, 3), (6, 2)}

∴ n(D) = 4

$\therefore P ( D )=\frac{n(D)}{n(S)}=\frac{4}{36}=\frac{1}{9}$

3 coins can be drawn from these 8 coins in ${ }^8 C_3$ ways.

$\therefore n ( s )={ }^8 C _3=\frac{8 !}{5 ! 3 !}=\frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 3 \times 2 \times 1}=56$

B.5 letters have to be placed in 5 envelopes in such a way that no envelope is empty.

∴ The first letter can be placed into 5 envelopes in 5 different ways, the second letter in 4 ways.

Similarly, the third, fourth and fifth letters can be placed in 3 ways, 2 ways and 1 way, respectively.

∴ Total number of ways = 5! = 5 × 4 × 3 × 2 × 1 = 120

∴ n(S) = 120

C.6 books can be arranged on a shelf in ${ }^6 P_6=6$ ! ways.

∴ n(S) = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

D.3 tickets are drawn at random from 20 tickets.

$\therefore 3$ tickets can be selected in ${ }^{20} C _3$ ways.

$\therefore n ( S )={ }^{20} C _3=\frac{20 !}{17 ! 3 !}=\frac{20 \times 19 \times 18 \times 17 !}{17 ! \times 3 \times 2 \times 1}=1140$