Question

A problem in calculus is given to 2 students A and B, whose chances of solving it are $\dfrac{1}{3}$ and $\dfrac{1}{4}$ respectively. Find the probability of the problem being solved; if both of them try independently.

Answer 1

An event ‘A’ associated with a random experiment is said to occur if any one of the elementary events ‘A’ outcome of a random experiment is called an elementary event. Associated to the event is an outcome. If there are $n$ elementary events associated with a random experiment and $m$of them are favorable to an event A, then the probability of happening or occurrence of event A is denoted by $P(A)$ and is defined as the ratio $\dfrac{m}{n}$.
Thus, $P(A)$=$\dfrac{m}{n}$.
Formula to find probability when independent events ${E_1}$ and ${E_2}$ are given :$P({E_1} \cup {E_2}) = P({E_1}) + P({E_2}) - P({E_1} \cap {E_2})$, or
Formula for finding probability when A and B are given as independent events = $P(A) \times P(notB) + P(notA) \times P(B) + P(A) \times P(B)$.

Complete step-by-step answer:
Let ${E_1}$ and ${E_2}$ denote the events that the problem is solved by A and B respectively.
We have $P({E_1}) = \dfrac{1}{3}$and $P({E_2}) = \dfrac{1}{4}$
As these two are independent events, therefore
Required probability is: $P({E_1} \cup {E_2}) = P({E_1}) + P({E_2}) - P({E_1} \cap {E_2})$
As $P({E_1} \cap {E_2}) = P({E_1}) \times P({E_2})$
$\therefore$ $P({E_1} \cup {E_2}) = P({E_1}) + P({E_2}) - P({E_1}) \times P({E_2})$
Now substituting $P({E_1}) = \dfrac{1}{3}$and$P({E_2}) = \dfrac{1}{4}$, we get
$P({E_1} \cup {E_2}) = \dfrac{1}{3} + \dfrac{1}{4} - \left( {\dfrac{1}{3} \times \dfrac{1}{4}} \right)$
By solving the bracket first, we get
$P({E_1} \cup {E_2}) = \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{{12}}$
Taking L.C.M of denominators and solving it, we get
$\Rightarrow P({E_1} \cup {E_2}) = \dfrac{1}{2}$
$\because {E_1}$ and ${E_2}$ are independent
$\therefore P({E_1} \cup {E_2}) = \dfrac{1}{2}$

$\therefore$ The probability of the problem being solved; if both of them try independently= $\dfrac{1}{2}$.

Note: Alternative method: the probability that the problem is solved will be found by finding the probability that either A solves or B solves or A and B solve the problem.
Here $P(A) = \dfrac{1}{3}$ and $P(B) = \dfrac{1}{4}$.
Probability of problem being solved=$P(A) \times P(\overline {B)} + P(\overline {A)} \times P(B) + P(A) \times P(B)$
Substituting the values of $P(A) = \dfrac{1}{3}$,$P(B) = \dfrac{1}{4}$, $P(\overline A ) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$ and $P(\overline B ) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$.
We have,
Probability of problem being solved= $\dfrac{1}{3} \times \dfrac{3}{4} + \dfrac{2}{3} \times \dfrac{1}{4} + \dfrac{1}{3} \times \dfrac{1}{4}$
$\therefore$ Probability of problem being solved= $\dfrac{6}{{12}} = \dfrac{1}{2}$
$\therefore$The probability of the problem being solved; if both of them try independently=$\dfrac{1}{2}$.