Question

A bag has $13$ red, $14$ green and $15$ white balls. $p_1$ is the probability of drawing exactly $2$ white balls when four balls are drawn. Then the number of balls of each color are doubled. Let $p_2$ be the probability of drawing $4$ balls when $8$ balls are drawn, then

• A. $p_1=p_2$
• B. $p_1>p_2$
• C. $p_1<p_2$
• D. none of these

Answer 1

Answer: (B)

$p_1>p_2$

$p_1=\frac{{}^{15}{C}_2}{{}^{42}{C}_4}=\frac{15\times 14\times 4!}{2!\times 42\times 41\times 40\times 39}=\frac{1}{41\times 26}$ and

$p_2=\frac{{}^{30}{C}_4}{{}^{84}{C}_8}=\frac{15\times 14\times 13\times 12\times 8!}{4!\times 84\times 83\times 82\times ...\times 77}$

$=\frac{15\times 14\times 13\times 12\times 8\times 7\times 6\times 5}{84\times 83\times ...\times 77}<p_2\Rightarrow p_1>p_2$