Question

Out of 13 applicants for a job, there are 5 women and 8 men. It is desired to select 2 persons for the job. The probability that at least one of the selected persons is a woman is:
[a] $\dfrac{25}{39}$
[b] $\dfrac{14}{39}$
[c] $\dfrac{5}{13}$
[d] $\dfrac{10}{13}$

Answer 1

Hint: Probability of event E = $\dfrac{n(E)}{n(S)}$= $\dfrac{\text{Favourable cases}}{\text{Total number of cases}}$ where S is called the sample space of the random experiment. Find n (E) and n (S) and use the above formula to find the probability. Recall that the number of ways of choosing r persons among n persons is given by $^{n}{{C}_{r}}$. Find the number of ways in which 2 persons can be selected among 5 women and 8 men so that no women is selected. Use complementary principles to find the number of ways in which 2 persons can be selected from 5 women and 8 men so that at least one woman is selected. Find the total number of ways of selecting 2 applicants out of 13 applicants without any restriction. Hence use the above formula to find the probability.

Complete step-by-step answer:

Let E be the event that among the selected persons, at least one is women.
Then we have
E’: Both of the selected persons are men.
n(E’) = number of ways in which 2 men can be selected out of 5 women and 8 men = number of ways in which 2 men can be selected out of 8 men ${{=}^{8}}{{C}_{2}}$ .
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Using we get
n(E’) $=\dfrac{8!}{2!\left( 8-2 \right)!}=\dfrac{8\times 7\times 6!}{2!6!}=\dfrac{56}{2}=28$
Also n(S) = number of ways in which 2 persons can be selected out of 13 persons ${{=}^{13}}{{C}_{2}}=\dfrac{13!}{2!11!}=\dfrac{13\times 12\times 11!}{2!11!}=\dfrac{156}{2}=78$
Hence $\text{P(E }\!\!'\!\!\text{ )=}\dfrac{n(E')}{n(S)}=\dfrac{28}{78}=\dfrac{14}{39}$
Now, we know that $\text{P(E)=1-P(E }\!\!'\!\!\text{ )}$
Using, we get
$\text{P(E)=1-}\dfrac{14}{39}=\dfrac{25}{39}$.
Hence the probability that at least one of the selected persons is a woman is $\dfrac{25}{39}$ .
Hence option [a] is correct.

Note: [1] The probability of an event always lies between 0 and 1
[2] The sum of probabilities of an event E and its complement E’ = 1
i.e. $P(E)+P(E')=1$
Hence, we have $P(E')=1-P(E)$. This formula is applied when it is easier to calculate P(E’) instead of P(E).
In the question, we have F = E’. Hence $P(F)=1-P(E)=1-\dfrac{1}{2}=\dfrac{1}{2}$ which is the same as above.
[3] Probability is the likelihood of an event and not the guarantee. It means that if the probability of an event is 0.5, it does not mean that the event occurs half the time. It only ascertains that the likelihood of occurrence of the event is half the times.