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. None of these

Answer 1

The correct option is C

$p_1>p_2$

Explanation for the correct option:

Step-1:The probability of getting exactly $2$ blacks on pulling out $4$ balls:

A bag has $13$ red, $14$ green and $15$ black balls

Total balls $=42$

According to the question,

$p_1=\frac{{}^{15}C_2.{}^{27}C_2}{{}^{42}C_4}$

$=\frac{15\times 14\times 27\times 26\times 1\times 2\times 3\times 4}{1\times 2\times 1\times 2\times 42\times 41\times 40\times 39}$

$=\frac{27}{82}$

Step-2:The probability of getting exactly $2$ blacks on pulling out $8$ balls:

$p_2=\frac{{}^{30}C_4.{}^{54}C_4}{{}^{84}C_8}$

$=\frac{30\times 29\times 28\times 27\times 54\times 53\times 52\times 51\times 8!}{4!\times 4!\times 84\times 83\times 82\times 81\times 80\times 79\times 78\times 77}$

$=\frac{17\times 28\times 45\times 53}{11\times 79\times 82\times 83}$

$\therefore$$\frac{p_1}{p_2}=\frac{27}{82}\times \frac{11\times 79\times 82\times 83}{17\times 29\times 45\times 53}$

$=\frac{33\times 79\times 83}{29\times 53\times 85}$

$=\frac{216381}{130645}>1$

Thus, $p_1>p_2$

Hence, option ‘C’ is correct.