Maths Solve the Following Question.(5 Marks)

Since A, B are disjoint,

A ∩ B = Φ and A ∩ B ∩ C = Φ

∴ P(A ∩ B) = 0, P(A ∩ B ∩ C) = 0 ……(i)

Since A and C are independent,

P(A ∩ C) = P(A) P(C) = xz

Since B and C are independent,

P(B ∩ C) = P(B) P(C) = yz

P(A ∪ C) = P(A) + P(C) – P(A ∩ C)

$\therefore \frac{2}{3}=X+Z-X Z \ldots \ldots \ldots$

..(ii)

P(B ∪ C) = P(B) + P(C) – P(B ∩ C)

$\therefore \frac{3}{4}=y+z-y z$.

……(iii)

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)

$\begin{aligned} & \frac{11}{12}=x+y+z-0-y z-z x+0 \ldots \ldots[\text { From(i)] } \\ & =(x+z-x z)+(y+z-y z)-z \\ & =\frac{2}{3}+\frac{3}{4}-z \ldots \ldots .[\text { From (ii) and (iii)] }\end{aligned}$

$\therefore \quad z =\frac{2}{3}+\frac{3}{4}-\frac{11}{12}=\frac{17-11}{12}=\frac{1}{2}$

Substituting $z=\frac{1}{2}$ in (ii), we get

$\begin{aligned} \frac{2}{3} & =x+\frac{1}{2}-\frac{1}{2} x \\ \therefore \quad x & =2\left(\frac{2}{3}-\frac{1}{2}\right)=\frac{2}{6}\end{aligned}$

$\therefore \quad x=\frac{1}{3}$

Substituting $z=\frac{1}{2}$ in (iii), we get

$\begin{array}{ll} & \frac{3}{4}=y+\frac{1}{2}-\frac{1}{2} y \\ \therefore \quad & y=2\left(\frac{3}{4}-\frac{1}{2}\right)=2\left(\frac{1}{4}\right)=\frac{1}{2} \\ \therefore \quad & P ( A )=\frac{1}{3}, P ( B )=\frac{1}{2}, P ( C )=\frac{1}{2}\end{array}$

$P(A)=\frac{1}{5}$

$\begin{aligned} & \text { and } \frac{P(A \cap B)}{P(B)}=\frac{1}{5} \\ & \therefore P ( A )=\frac{1}{5} \ldots \ldots \text { (i) } \\ & P ( B )=5 P ( A \cap B ) \ldots \ldots . \text { (ii) } \\ & \text { Since } P ( B / A )=\frac{1}{3} \\ & \frac{P(A \cap B)}{P(A)}=\frac{1}{3} \\ & \therefore P ( A )=3 P ( A \cap B ) \ldots \ldots . . \text { (iii) }\end{aligned}$

$\begin{aligned} P\left(A^{\prime} / B\right) & =\frac{P\left(A^{\prime} \cap B\right)}{P(B)} \\ & =\frac{P(B)-P(A \cap B)}{P(B)} \\ & =1-\frac{P(A \cap B)}{P(B)}\end{aligned}$

$=1-\frac{1}{5} \quad \ldots[$ From (ii)]

$=\frac{4}{5}$

2.$3 P ( A \cap B )= P ( A ) \quad \ldots[$ From (iii) $]$

$\therefore \quad P ( A \cap B )=\frac{1}{3}\left(\frac{1}{5}\right)=\frac{1}{15}$

$\therefore \quad P ( B )=5 P ( A \cap B ) \quad \ldots[$ From (ii)]

$\begin{aligned}=\frac{5}{15} & =\frac{1}{3} \\ P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ & =\frac{1}{5}+\frac{1}{3}-\frac{1}{15}\end{aligned}$

$\begin{aligned} & \quad=\frac{3+5-1}{15}=\frac{7}{15} \\ & P\left(A^{\prime} \cap B^{\prime}\right)= P ( A \cup B )^{\prime} \\ & =1- P ( A \cup B )\end{aligned}$

$=1-\frac{7}{15}=\frac{8}{15}$

$\begin{aligned} P \left( B ^{\prime} / A ^{\prime}\right)=\frac{ P \left( A ^{\prime} \cap B ^{\prime}\right)}{ P \left( A ^{\prime}\right)} & =\frac{\frac{8}{15}}{1-\frac{1}{5}} \\ & =\frac{\frac{8}{15}}{\frac{4}{5}}=\frac{2}{3}\end{aligned}$

iii. $\begin{aligned} P\left(B^{\prime} / A\right) & =\frac{P\left(B^{\prime} \cap A\right)}{P(A)} \\ & =\frac{P\left(B^{\prime}\right) \times P(A)}{P(A)} \\ & =P\left(B^{\prime}\right) \\ & =1-P(B) \\ & =1-\frac{2}{5}=\frac{3}{5}\end{aligned}$

$\therefore P(A)=\frac{5}{12}$

Let event B: The wife will be alive till 65.

$\begin{aligned} & \therefore P(B)=\frac{3}{8} \\ & \therefore P\left(A^{\prime}\right)=1-P(A)=1-\frac{5}{12}=\frac{7}{12} \\ & P\left(B^{\prime}\right)=1-P(B)=1-\frac{3}{8}=\frac{5}{8}\end{aligned}$

Since A and B are independent events, A’ and B’ are also independent events.

(a) Let event C : Both man and his wife will be alive.

∴ P(C) = P(A ∩ B) = P(A) . P(B)

$\begin{aligned} & =\frac{5}{12} \times \frac{3}{8} \\ & =\frac{5}{32}\end{aligned}$

Let event D: Exactly one of them will be alive.

∴ P(D) = P(A’ ∩ B) + P(A ∩ B’)

= P(A’) . P(B) + P(A) . P(B’)

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

Let event E: None of them will be alive.

∴ P(E) = P(A’ ∩ B’) = P(A’) . P(B’)

$\begin{aligned} & =\frac{7}{12} \times \frac{5}{8} \\ & =\frac{35}{96}\end{aligned}$

Let event F: At least one of them will be alive.

∴ P(F) = 1 – P(none of them will be alive)

$\begin{aligned} & =1-\frac{35}{96} \\ & =\frac{61}{96}\end{aligned}$

event B: The patient had throat surgery.

Given, n(S) = 200

n(A ∩ B) = 70

$\begin{aligned} & \therefore P ( A \cap B )=\frac{ n ( A n B )}{ n ( S )}=\frac{70}{200} \\ & n ( B )=95 \\ & \therefore P ( B )=\frac{n(B)}{n(S)}=\frac{95}{200}\end{aligned}$

∴ Required probability = P(A / B)

$\begin{aligned} & =\frac{P(A \cap B)}{P(B)} \\ & =\frac{\left(\frac{70}{200}\right)}{\left(\frac{95}{200}\right)} \\ & =\frac{70}{95} \\ & =\frac{14}{19}\end{aligned}$

Check:

Reduce the sample space to the set of throat patients only.

n(S) = 95 Let E : Patient had satisfactory throat surgery.

n(E) = 70

$\therefore P ( E )=\frac{n(E)}{n(S)}=\frac{70}{95}=\frac{14}{19}$

Let event C : The patient was unsatisfied,

event D : The patient had a eye surgery.

Given, n(S) = 200 n(C ∩ D) = 15

$\begin{aligned} & \therefore P (C \cap D )=\frac{n(C \cap D)}{n(S)}=\frac{15}{200} \\ & n ( D )=105 \\ & \therefore P ( D )=\frac{105}{200}\end{aligned}$

Required probability = P(C / D)

$\begin{aligned} & =\frac{P(C \cap D)}{P(D)} \\ & =\frac{\left(\frac{15}{200}\right)}{\left(\frac{105}{200}\right)} \\ & =\frac{1}{7}\end{aligned}$

Let event F : The patient had a throat surgery,

event G : The patient was unsatisfied.

Given, n(S) = 200

n(F ∩ G) = 25

$\begin{aligned} & \therefore P ( F \cap G )=\frac{n(F \cap G)}{n(S)}=\frac{25}{200} \\ & n (G)=40 \\ & \therefore P (G)=\frac{n(G)}{n(S)}=\frac{40}{200}\end{aligned}$

∴ Required probability = P(F / G)

$\begin{aligned} & =\frac{P(F \cap G)}{P(G)} \\ & =\frac{\left(\frac{25}{200}\right)}{\left(\frac{40}{200}\right)} \\ & =\frac{5}{8}\end{aligned}$

$\begin{aligned} & \therefore P(A)=\frac{2}{5}, P(B)=\frac{1}{4} \\ & \therefore P\left(A^{\prime}\right)=1-P(A)=1-\frac{2}{5}=\frac{3}{5} \\ & P\left(B^{\prime}\right)=1-P(B)=1-\frac{1}{4}=\frac{3}{4}\end{aligned}$

Since A and B are independent events, A’ and B’ are also independent events.

(i) Let event C: Problem is not solved.

$\begin{aligned} & \therefore P(C)=P\left(A^{\prime} \cap B^{\prime}\right) \\ & =P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right) \\ & =\frac{3}{5} \times \frac{3}{4} \\ & =\frac{9}{20}\end{aligned}$

Let event b: Problem is solved.

Problem can be solved if at least one of the two students solves the problem.

∴ P(D) = P(at least one student solves the problem)

= 1 – P(no student solves the problem)

= 1 – P(A’ ∩ B’)

= 1 – P(A’) P(B’)

$\begin{aligned} & =1-\frac{3}{5} \times \frac{3}{4} \\ & =1-\frac{9}{20} \\ & =\frac{11}{20}\end{aligned}$

Let event c: The problem is solved exactly by one of them.

∴ P(E) = P(A’ ∩ B) ∪ P(A ∩ B’)

= P(A’) . P(B) + P(A) . P(B’)

$\begin{aligned} & =\left(\frac{3}{5} \times \frac{1}{4}\right)+\left(\frac{2}{5} \times \frac{3}{4}\right) \\ & =\frac{3}{20}+\frac{6}{20} \\ & =\frac{9}{20}\end{aligned}$

∴ n(S) = 75

(i) Let event A: The number on the ticket is a perfect square.

∴ A = {1, 4, 9, 16, 25, 36, 49, 64}

∴ n(A) = 8

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

Let event B: The number on the ticket is divisible by 4.

∴ B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72}

∴ n(B) = 18

$\therefore P ( B )=\frac{n(B)}{n(S)}=\frac{18}{75}$

Now, A ∩ B = {4, 16, 36, 64}

∴ n(A ∩ B) = 4

$\therefore P(A \cap B)=\frac{n(A \mid B)}{n(S)}=\frac{4}{75}$

∴ Required probability

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

$\begin{aligned} & =\frac{8}{75}+\frac{18}{75}-\frac{4}{75} \\ & =\frac{22}{75}\end{aligned}$

(ii)Let event A: The number on the ticket is a prime number.

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

∴ n(A) = 21

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{21}{75}$

Let event B: The number is greater than 40.

∴ B = {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75}

∴ n(B) = 35

$\therefore P ( B )=\frac{ n ( B )}{ n ( S )}=\frac{35}{75}$

Now,

A ∩ B = {41, 43, 47, 53, 59, 61, 67, 71, 73}

∴ n(A ∩ B) = 9

$\therefore n ( A \cap B )=\frac{ n ( A \cap B )}{ n ( S )}=\frac{9}{75}$

∴ Required probability

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

$\begin{aligned} & =\frac{21}{75}+\frac{35}{75}-\frac{9}{75} \\ & =\frac{47}{75}\end{aligned}$

∴ n(S) = 52

The pack of 52 cards consists of 26 red and 26 black cards.

(i) Let event A: A red card is drawn.

$\therefore$ Red card can be drawn in ${ }^{26} C _1=26$ ways

∴ n(A) = 26

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{26}{52}$

Let event B: A black card is drawn.

$\therefore$ Black card can be drawn in ${ }^{26} C _1=26$ ways.

∴ n(B) = 26

$\therefore P ( B )=\frac{n(B)}{n(S)}=\frac{26}{52}$

Since A and B are mutually exclusive events,

P(A ∩ B) = 0

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

$\begin{aligned} & =\frac{26}{52}+\frac{26}{52} \\ & =1\end{aligned}$

Let event A: A black card is drawn.

$\therefore$ Black card can be drawn in ${ }^{26} C _1=26$ ways.

n(A) = 26 n(A) 26 n(S) ~ 52

Let event B: A face card is drawn.

There are 12 face cards in the pack of 52 cards.

$\therefore 1$ face card can be drawn in ${ }^{12} C _1=12$ ways.

∴ n(B) = 12

$\therefore P(B)=\frac{n(B)}{n(S)}=\frac{12}{52}$

There are 6 black face cards. ∴ n(A ∩ B) = 6

$\therefore P ( A \cap B )=\frac{n(A \cap B)}{n(S)}=\frac{6}{52}$

∴ Required probability

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

$\begin{aligned} & =\frac{26}{52}+\frac{12}{52}-\frac{6}{52} \\ & =\frac{32}{52} \\ & =\frac{8}{13}\end{aligned}$

(b)one is a club and the other is a diamond.

(c)both are from the same suit.

(d)both are red ca

Two cards can be drawn from a pack of 52 cards in ${ }^{52} C _2$ ways.

$\therefore n(S)={ }^{52} C _2$

Let event A: Out of the two cards drawn, one is a face card and the other is an ace card. There are 12 face cards and 4 ace cards in a pack of 52 cards.

$\therefore$ One face card can be drawn from 12 face cards in ${ }^{12} C _1$ ways and one ace card can be

drawn from 4 ace cards in ${ }^4 C _1$ ways.

$\begin{aligned} & \therefore n ( A )={ }^{12} C _1 \times{ }^4 C _1 \\ & \therefore P ( A )=\frac{ n ( A )}{ n ( S )}=\frac{{ }^{12} C _1 \times{ }^4 C _1}{{ }^{52} C _2}=\frac{12 \times 4}{\frac{52 \times 51}{2 \times 1}}=\frac{8}{221}\end{aligned}$

Let event B: Out of the two cards drawn, one is club and the other is a diamond card.There are 13 club cards and 13 diamond cards.

$\therefore$ One club card can be drawn from 13 club cards in ${ }^{13} C _1$ ways and one diamond card can

be drawn from 13 diamond cards in ${ }^{13} C _1$ ways.

$\begin{aligned} & \therefore n ( B )={ }^{13} C_1 \times{ }^{13} C_1 \\ & \therefore P ( B )=\frac{n(B)}{n(S)}=\frac{{ }^{13} C_1 \times{ }^{13} C_1}{{ }^{52} C_2}=\frac{13 \times 13}{\frac{52 \times 51}{2 \times 1}}=\frac{13}{102}\end{aligned}$

Let event C: Both the cards drawn are of the same suit.A pack of 52 cards consists of 4 suits each containing 13 cards.

$\therefore 2$ cards can be drawn from the same suit in ${ }^{13} C _2$ ways.

$\begin{aligned} & \therefore n ( C )={ }^{13} C _2 \times 4 \\ & \therefore P ( C )=\frac{ n ( C )}{ n ( S )}=\frac{4 \times{ }^{13} C _2}{{ }^{52} C _2}=\frac{4 \times 13 \times 12}{52 \times 51}=\frac{4}{17}\end{aligned}$

Let event D: Both the cards drawn are red. There are 26 red cards in the pack of 52 cards.

$\therefore 2$ cards can be drawn from them in ${ }^{26} C _2$ ways.

$\begin{aligned} & \therefore n ( D )={ }^{26} C _2 \\ & \therefore P ( D )=\frac{n(D)}{n(S)}=\frac{{ }^{26} C_2}{{ }^{52} C_2}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\end{aligned}$

Let event E: Out of the two cards drawn, one is heart and other is non-heart. There are 13 heart cards in a pack of 52 cards, i.e., 39 cards are non-heart.

$\therefore$ One heart card can be drawn from 13 hdart cards in ${ }^{13} C _1$ ways and one non-heart card

can be drawn from 39 cards in ${ }^{39} C_1$ ways.

$\begin{aligned} & \therefore n ( E )={ }^{13} C_1\times{ }^{39} C _1 \\ &\therefore P ( E )=\frac{ n ( E )}{ n ( S )}=\frac{{ }^{13} C _1 \times{ }^{39} C _1}{{ }^{52} C _2}=\frac{13\times 39}{\frac{52 \times 51}{2 \times 1}}=\frac{13}{34}\end{aligned}$