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 colour are doubled. Let $p_2$ be the probability of drawing 4 white balls when 8 ball 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)

$p_1=\frac{{}^{15}{C}_2}{{}^{42}{C}_4}=\frac{15\times 14\times 4!}{2!\times 42\times 41\times 40\times 39}=\frac{1}{41\times 26}$ and
$P_2=\frac{30_{C_4}}{{}^{84}{C}_8}=\frac{15\times 14\times 13\times 12\times 8!}{4!\times 84\times 83\times 82\times .....\times 77}$
$=\frac{15\times 14\times 13\times 12\times 8\times 7\times 6\times 5}{84\times 83\times 82\times 81\times 79\times 78\times 77}<p_1\Rightarrow p_1>p_2.$