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 will be a woman is

• A. $\frac{25}{39}$
• B. $\frac{14}{39}$
• C. $\frac{5}{13}$
• D. $\frac{10}{13}$

Answer 1

The correct option is A $\frac{25}{39}$

Number of ways in which$2$ people are chosen from $13$ people is
${{13}_C}_2=\frac{13!}{2!(13-2)!}=\frac{13!}{2!\left(11\right)!}=\frac{13\times 12\times 11!}{2\times 1\times 11!}=78$
For atleast one person selected to be a women,
Case 1 : When both persons selected are women.
Number of ways in which$2$ women be chosen from $5$ women is
${5_C}_2=\frac{5!}{2!(5-2)!}=\frac{5!}{2!\left(3\right)!}=\frac{5\times 4\times 3!}{2\times 1\times 3!}=10$
Here we applied the formula ${n_C}_r=\frac{n!}{r!(n-r)!}$
Case 2: When one man is selected and one woman is selected
Number of ways in which$1$ man can be chosen from $8$ men and $1$ woman can be chosen from $5$ women is
${8_C}_1\times {5_C}_1=8\times 5=40$
So, Probability $=\frac{NumberoffavourableOutcomes}{Totalnumberofoutcomes}=\frac{10+40}{78}=\frac{25}{39}$