Question

A bag has $13$ red, $14$ green and $15$ black balls. The probability of getting exactly $2$ blacks on pulling out $4$ balls is $p_1.$ Now the number of each colour ball is doubled and 8 balls are pulled out. The probability of getting exactly $4$ blacks is $p_2.$ Then

• A. $P_1=P_2.$
• B. $P_1>P_2$
• C. $P_1<P_2.$
• D. Data Insufficient

Answer 1

The correct option is B $P_1>P_2$

$P_1=\frac{{}^{15}{C}_2{\times }^{27}{C}_2}{{}^{42}{C}_4}$ $=\frac{15\times 14\times 27\times 26\times 1\times 2\times 3\times 4}{1\times 2\times 1\times 2\times 42\times 41\times 40\times 39}=\frac{27}{82}.$

$p_2=\frac{{}^{30}{C}_4{\times }^{54}{C}_4}{{}^{84}{C}_8}$ $=\frac{\mathrm{30.29.28.27.54.53.52.51.8}!}{4!.4!.84.83.82.81.80.79.78.77}$ $=\frac{\mathrm{17.28.45.53}}{\mathrm{11.79.82.83}},$ after simplification.

$\therefore \frac{p_1}{p_2}=\frac{27}{82}\times \frac{\mathrm{11.79.82.83.}}{\mathrm{17.29.45.53}}=\frac{\mathrm{33.79.83}}{\mathrm{29.53.85}}=\frac{216381}{130645}>1.$ Hence $p_1>p_2.$