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
1. ${p_1} < {p_2}$
2. ${p_1} = {p_2}$
3. ${p_1} > {p_2}$
4. None of these

Answer 1

Hint: In this question, we are given two probabilities ${p_1}$, which is the probability of getting exactly $2$ blacks on pulling out $4$ balls and ${p_2}$ is the probability of getting exactly $4$ blacks on pulling out $8$balls where the number of balls is doubled. We have to find the relation between both probabilities. The first step is to find the value of ${p_1}$ using a combination formula by pulling $2$ balls from the black and other $2$ from the total of remaining balls. Now double the number of each color ball and apply the same formula to find the probability. In the last divide both the probabilities by each other.

Formula Used:
Combination formula –
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ........$

Complete step by step Solution:
Given that,
${p_1}$ is the probability of getting exactly $2$ black balls on pulling $4$ balls
And similarly, ${p_2}$ is the probability of getting exactly $4$ black balls on pulling $8$ balls when each color ball is doubled.
Case 1:
Number of red, green, and black balls are $13$, $14$, $15$ respectively
Total number of balls $ = 42$
Now, using combination formula
${p_1} = \dfrac{{{}^{15}{C_2} \times {}^{27}{C_2}}}{{{}^{42}{C_4}}}$
$ = \dfrac{{\dfrac{{15!}}{{2!\left( {15 - 2} \right)!}} \times \dfrac{{27!}}{{2!\left( {27 - 2} \right)!}}}}{{\dfrac{{42!}}{{4!\left( {42 - 4} \right)!}}}}$
$ = \dfrac{{\dfrac{{15!}}{{2!13!}} \times \dfrac{{27!}}{{2!25!}}}}{{\dfrac{{42!}}{{4!38!}}}}$
$ = \dfrac{{\dfrac{{15 \times 14 \times 13!}}{{2 \times 13!}} \times \dfrac{{27 \times 26 \times 25!}}{{2 \times 25!}}}}{{\dfrac{{42 \times 41 \times 40 \times 39 \times 38!}}{{4 \times 3 \times 2 \times 1 \times 38!}}}}$
$ = \dfrac{{15 \times 14 \times 27 \times 26 \times 4 \times 3 \times 2}}{{2 \times 2 \times 42 \times 41 \times 40 \times 39}}$
$ = \dfrac{{27}}{{82}}$
Case 2:
When number of each colour ball is doubled.
Number of red, green, and black balls are $26$, $28$, $30$ respectively
Total number of balls $ = 84$
Again, using combination formula
${p_2} = \dfrac{{{}^{30}{C_4} \times {}^{54}{C_4}}}{{{}^{84}{C_8}}}$
$ = \dfrac{{\dfrac{{30!}}{{4!\left( {30 - 4} \right)!}} \times \dfrac{{54!}}{{4!\left( {54 - 4} \right)!}}}}{{\dfrac{{84!}}{{8!\left( {84 - 8} \right)!}}}}$
$ = \dfrac{{\dfrac{{30!}}{{4!26!}} \times \dfrac{{54!}}{{4!50!}}}}{{\dfrac{{84!}}{{8!76!}}}}$
$ = \dfrac{{\dfrac{{30 \times 29 \times 28 \times 27 \times 26!}}{{4 \times 3 \times 2 \times 1 \times 26!}} \times \dfrac{{54 \times 53 \times 52 \times 51 \times 50!}}{{4 \times 3 \times 2 \times 1 \times 50!}}}}{{\dfrac{{84 \times 83 \times 82 \times 81 \times 80 \times 79 \times 78 \times 77 \times 76!}}{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 76!}}}}$
$ = \dfrac{{30 \times 29 \times 28 \times 27 \times 54 \times 53 \times 52 \times 51 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}}{{4 \times 3 \times 2 \times 4 \times 3 \times 2 \times 84 \times 83 \times 82 \times 81 \times 80 \times 79 \times 78 \times 77}}$
$ = \dfrac{{1135260}}{{5914414}}$
Now, dividing ${p_1}$by ${p_2}$,
$\dfrac{{{p_1}}}{{{p_2}}} = \dfrac{{\left( {\dfrac{{27}}{{82}}} \right)}}{{\left( {\dfrac{{1135260}}{{5914414}}} \right)}}$
$ = \dfrac{{27 \times 5914414}}{{82 \times 1135260}}$
$ = \dfrac{{216381}}{{130645}}$
$ = 1.66 > 1$
Since $\dfrac{{{p_1}}}{{{p_2}}} > 1$
It implies that, ${p_1} > {p_2}$

Hence, the correct option is 3.

Note: The key concept involved in solving this problem is the good knowledge of combination and permutation in probability. Students must remember that a combination is a mathematical technique for determining the number of possible arrangements in a set of items where the order of the selection is irrelevant. You can choose the items in any order in combinations. Permutations and combinations are often confused. In permutations, however, the order of the selected items is critical.