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)

$p_{1}=\frac{^{15}C_{2}}{^{42}C_{4}}=\frac{15\times14\times4!}{2!\times42\times41\times40\times39}=\frac{1}{41\times26}$ and
$P_{2}=\frac{30_{C_4}}{^{84}C_{8}}=\frac{15\times14\times13\times12\times8!}{4!\times84\times83\times82\times .....\times77}$
$=\frac{15\times14\times13\times 12\times8\times7\times6\times5}{84\times83\times82\times81\times79\times78\times77} < p_{1} \Rightarrow p_{1} > p_{2}.$