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 exactly one of them problem

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Probability of solving the problem by $\mathrm{A},\, \mathrm{P}(\mathrm{A})=\frac{1}{2}$ 

Probability of solving the problem by $\mathrm{B}, \,\mathrm{P}(\mathrm{B})=\frac{1}{3}$ 

since the problem is solved independently by $A$ and $B$,

$\therefore $ $\mathrm{P}(\mathrm{AB})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$

$P(A^{\prime})=1-P(A)=1-\frac{1}{2}=\frac{1}{2}$

$P(B^{\prime})=1-P(B)=1-\frac{1}{3}=\frac{2}{3}$

Probability that exactly one of them solves the problem is given by,

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

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

$=\frac{1}{3}+\frac{1}{6}$

$=\frac{1}{2}$

Similar Questions

For two given events $A$ and $B$, $P\,(A \cap B) = $

  • [IIT 1988]

If $A$ and $B$ are two events such that $P\left( {A \cup B} \right) = P\left( {A \cap B} \right)$, then the incorrect statement amongst the following statements is

  • [JEE MAIN 2014]

Fill in the blanks in following table :

$P(A)$ $P(B)$ $P(A \cap B)$ $P (A \cup B)$
$0.5$ $0.35$ .........  $0.7$

Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined $P ( A )=0.5$,  $ P ( B )=0.4$,  $P ( A \cap B )=0.8$

A card is drawn at random from a pack of cards. The probability of this card being a red or a queen is