Question

If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?
A. 24/91
B. 45/91
C. 2/3
D. 67/91
E. 84/91

Answer 1

Answer:

D. $\frac{67}{91}$

Step-by-step explanation:

Given,

Total number of jurors = 15,

Out of which 2/3 are men and 1/3 are women,

So, total men = $\frac{2}{3}\times 15$ = 10,

And, total women = 15 - 10 = 5,

The people have to select = 12,

So, the total ways of selecting 12 people = $^{15}C_{12}=\frac{15!}{12!\times 3!}=455$

Now, 2/3 of 12 = $\frac{2}{3}\times 12$  = 2 × 4 = 8,

Thus, there would be at least 8 men.

Now, the possible ways of selecting at least 8 men = 8 men 4 women + 9 men 3 women + 10 men 2 women

$=^{10}C_8\times ^5C_4 + ^{10}C_9\times ^5C_3+^{10}C_{10}\times ^5C_2$

$=\frac{10!}{8!\times 2!}\times \frac{5!}{4!\times 1!}+\frac{10!}{9!\times 1!}\times \frac{5!}{3!\times 2!}+\frac{10!}{10!\times 0!}\times \frac{5!}{2!\times 3!}$

= 225 + 100 + 10

= 335

Hence, the probability that the jury will comprise at least 2/3 men

$=\frac{\text{the possible ways of selecting at least 8 men}}{\text{total ways of selecting 12 people}}$

$=\frac{335}{455}$

$=\frac{67}{91}$

OPTION D is correct.