Question

Three persons work independently on a problem. If the respective probabilities that they will solve it are $\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5}$, then find the probability that none can solve it?
(a) $\dfrac{3}{5}$
(b) $\dfrac{2}{5}$
(c) $\dfrac{1}{5}$
(d) 0

Answer 1

It is given that the three persons are doing a problem independently and also the respective probabilities of the three persons are also given. We are asked to find the probability that none can solve it which we are going to calculate by multiplying the probabilities when each of the three persons cannot work on the problem. We know the probabilities when each of the three persons can work on the problem so when we subtract 1 from each of the three probabilities, we will get the probability when none can solve the problem.

Complete step-by-step answer:
The probabilities when three persons work independently on the problem are:
$\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5}$
Let us assume that ${{E}_{1}},{{E}_{2}},{{E}_{3}}$ are three events when the three persons can solve the problem and their respective probabilities are:
$\begin{aligned} & P\left( {{E}_{1}} \right)=\dfrac{1}{3} \\\ & P\left( {{E}_{2}} \right)=\dfrac{1}{4} \\\ \end{aligned}$
$P\left( {{E}_{3}} \right)=\dfrac{1}{5}$
Now, the probability when events ${{E}_{1}},{{E}_{2}},{{E}_{3}}$ will not occur is equal to:
$\begin{aligned} & P\left( \overline{{{E}_{1}}} \right)=1-\dfrac{1}{3}=\dfrac{2}{3} \\\ & P\left( \overline{{{E}_{2}}} \right)=1-\dfrac{1}{4}=\dfrac{3}{4} \\\ & P\left( \overline{{{E}_{3}}} \right)=1-\dfrac{1}{5}=\dfrac{4}{5} \\\ \end{aligned}$
Now, multiplying the above three probabilities we get the probability when none of the three persons can solve the problem.
$\dfrac{2}{3}\left( \dfrac{3}{4} \right)\left( \dfrac{4}{5} \right)$
In the above expression, 3 and 4 will be cancelled out from the numerator and the denominator and we get,
$\dfrac{2}{5}$

So, the correct answer is “Option (b)”.

Note: You might have been thinking why we have multiplied the three probabilities, we can add them too. The answer is it is given that the three persons work independently on the problem it means all the three events (each event corresponds to a person’s probability of solving a problem) are independent to each other and we know that if we have three events ${{E}_{1}},{{E}_{2}},{{E}_{3}}$ independent to each other, then their collective probability is equal to multiplication of the probability of each event.
$P\left( {{E}_{1}} \right)P\left( {{E}_{2}} \right)P\left( {{E}_{3}} \right)$