Question

A) Probability of solving specific problem independently by $A$ and $B$ are $\frac{1}{2}\text{and}\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that
i) The problem is solved

B) Probability of solving specific problem independently by $A$ and $B$ are $\frac{1}{2}and\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that
ii) Exactly one of them solves the problem

Answer 1

A) Given:

Probability of solving the problem by $A$,

$P\left(A\right)=\frac{1}{2}$

Probability of solving the problem by $B$,

$P\left(B\right)=\frac{1}{3}$

Since, $A$ and $B$ are independent events

$P(A\cap B)=P\left(A\right)\times P\left(B\right)...\left(1\right)$

Substituting the value of $P\left(A\right)\&P\left(B\right)\text{in}\left(1\right)$,

$P(A\cap B)=\frac{1}{2}\times \frac{1}{3}$

$P(A\cap B)=\frac{1}{6}$

$P(A\cup B)$ = Probability that the problem is solved

We know that,

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)...\left(2\right)$

Substituting the value of $P\left(A\right),P\left(B\right)\&P(A\cap B)\text{in}\left(2\right)$,

$P(A\cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$\therefore P(A\cup B)=\frac{2}{3}$

Probability that the problem is solved $=\frac{2}{3}$

B) Given:

Probability of solving the problem by $A$,

$P\left(A\right)=\frac{1}{2}$

Probability of solving the problem by $B$,

$P\left(B\right)=\frac{1}{3}$

Since, $A$ and $B$ are independent events

$P(A\cap B)=P\left(A\right)\times P\left(B\right)...\left(1\right)$

Substituting the value of $P\left(A\right)\&P\left(B\right)\text{in}\left(1\right)$,

$P(A\cap B)=\frac{1}{2}\times \frac{1}{3}$

$P(A\cap B)=\frac{1}{6}$

Probability that exactly one of them solves the problem

= Probability that only $A$ solves + Probability that only $B$ solves

$=P(A\cap B^{\prime })+P(B\cap A^{\prime })$

$=\left(P\right(A)-P(A\cap B\left)\right)+\left(P\right(B)-P(B\cap A\left)\right)$

$P$(exactly one of them solves)

$=P\left(A\right)+P\left(B\right)-2P(A\cap B)...\left(2\right)$

Substituting the value of $P\left(A\right),P\left(B\right)\&P(A\cap B)\text{in}\left(2\right),$

$P$(exactly one of them solves)

$=\frac{1}{2}+\frac{1}{3}-2\times \frac{1}{6}$

$=\frac{1}{2}+\frac{1}{3}-\frac{1}{3}=\frac{1}{2}$

$\therefore \text{P(exactly one of them solves)}=\frac{1}{2}$