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Question: A bag contains a large number of white and black marbles in equal proportions. Two samples of 5 marb...

A bag contains a large number of white and black marbles in equal proportions. Two samples of 5 marbles are selected (with replacement) at random. Then the probability that the first sample contains exactly 1 black marble and the second sample contains exactly 3 black marbles is
a) 25512 b) 1332 c) 151024 d) −25256 \begin{aligned} & \text{a) }\dfrac{25}{512} \\\ & \text{b) }\dfrac{13}{32} \\\ & \text{c) }\dfrac{15}{1024} \\\ & \text{d) }\dfrac{-25}{256} \\\ \end{aligned}

Explanation

Solution

Let us assume there are n white and n black marbles. Now first we will calculate the total number of ways in which we can select 5 marbles.
Now for the first sample we will first select 1 black ball out on n black marbles and 4 white marbles out of n white marbles
Similarly for 2nd2^{nd} sample we will select 3 black marbles out of n black marbles and 2 white marbles out of n white marbles
Now we can calculate probability by formula number of required selectionnumber of total selection\dfrac{\text{number of required selection}}{\text{number of total selection}}

Complete step by step answer:
Now let us assume that there are 2n total marbles
Since it is given that there the white and black marbles are in equal proportions we can say that there are n black marbles and n white marbles.
Now sample of 5 marbles can be selected in 2nC5^{2n}{{C}_{5}} ways
Hence the total number of possible selections is 2nC5^{2n}{{C}_{5}}
Now let us first work for the first sample
Now in the first sample, we need exactly black marble. Now since we are selecting a total of 5 marbles we can say that rest 4 marbles will be white.
Now we know that there are total n black and n white marbles.
Hence the number of ways of selecting one black marble and 4 white marbles is nC1nC4^{n}{{C}_{1}}^{n}{{C}_{4}} …….. (2)
Now probability is number of required selectionnumber of total selection\dfrac{\text{number of required selection}}{\text{number of total selection}}
Hence from (1) and (2) we get.
Probability that 1st1^{st} sample has exactly 1 black marbles is nC1nC42nC5.................(3)\dfrac{^{n}{{C}_{1}}^{n}{{C}_{4}}}{^{2n}{{C}_{5}}}.................(3)
Now consider the second sample.
Second sample contains exactly three black marbles. Hence the rest two marbles are white
Now the number of ways of selecting 3 black marbles and 2 black marbles is nC3nC2................(4)^{n}{{C}_{3}}^{n}{{C}_{2}}................(4)
Now probability is number of required selectionnumber of total selection\dfrac{\text{number of required selection}}{\text{number of total selection}}
Hence from (1) and (4) we get.
Probability that 2nd2^{nd} sample has exactly 3 black marbles is nC3nC22nC5.......................(5)\dfrac{^{n}{{C}_{3}}^{n}{{C}_{2}}}{^{2n}{{C}_{5}}}.......................(5)
Hence from equation (4) and (5) we get
Required probability is equal to nC3nC22nC5×nC1nC42nC5\dfrac{^{n}{{C}_{3}}^{n}{{C}_{2}}}{^{2n}{{C}_{5}}}\times \dfrac{^{n}{{C}_{1}}^{n}{{C}_{4}}}{^{2n}{{C}_{5}}}
Now since n is given as very large number we will take limit n tending to infinity
Hence we get required probability is equal to lim⁡n→∞nC3nC22nC5×nC1nC42nC5{{\lim }_{n\to \infty }}\dfrac{^{n}{{C}_{3}}^{n}{{C}_{2}}}{^{2n}{{C}_{5}}}\times \dfrac{^{n}{{C}_{1}}^{n}{{C}_{4}}}{^{2n}{{C}_{5}}}
=lim⁡n→∞n!(3!)(n−3)!.n!(2!)(n−2)!(2n)!(2n−5)!5!×n!(n−1)!1!.n!(n−4)!4!(2n)!(2n−5)!5!={{\lim }_{n\to \infty }}\dfrac{\dfrac{n!}{(3!)(n-3)!}.\dfrac{n!}{(2!)(n-2)!}}{\dfrac{\left( 2n \right)!}{(2n-5)!5!}}\times \dfrac{\dfrac{n!}{(n-1)!1!}.\dfrac{n!}{(n-4)!4!}}{\dfrac{\left( 2n \right)!}{(2n-5)!5!}}
=lim⁡n→∞=n(n−1)(n−2)6.n(n−1)2.n1.n(n−1)(n−2)(n−3)24((2n)(2n−1)(2n−2)(2n−3)(2n−4))2120×120={{\lim }_{n\to \infty }}=\dfrac{\dfrac{n(n-1)(n-2)}{6}.\dfrac{n(n-1)}{2}.\dfrac{n}{1}.\dfrac{n(n-1)(n-2)(n-3)}{24}}{\dfrac{{{\left( \left( 2n \right)\left( 2n-1 \right)\left( 2n-2 \right)\left( 2n-3 \right)\left( 2n-4 \right) \right)}^{2}}}{120\times 120}}
=lim⁡n→∞14400288.n4(n−1)3(n−2)2(n−3)(2n)2(2n−1)2(2n−2)2(2n−3)2(2n−4)2={{\lim }_{n\to \infty }}\dfrac{14400}{288}.\dfrac{{{n}^{4}}{{(n-1)}^{3}}{{(n-2)}^{2}}(n-3)}{{{\left( 2n \right)}^{2}}{{\left( 2n-1 \right)}^{2}}{{\left( 2n-2 \right)}^{2}}{{\left( 2n-3 \right)}^{2}}{{\left( 2n-4 \right)}^{2}}}

& ={{\lim }_{n\to \infty }}\dfrac{14400}{288}.\dfrac{{{n}^{4}}{{(n-1)}^{3}}{{(n-2)}^{2}}(n-3)}{4{{n}^{2}}{{\left( 2n-1 \right)}^{2}}4{{\left( n-1 \right)}^{2}}{{\left( 2n-3 \right)}^{2}}4{{\left( n-2 \right)}^{2}}} \\\ & ={{\lim }_{n\to \infty }}\dfrac{225}{288}.\dfrac{{{n}^{2}}(n-1)(n-3)}{{{\left( 2n-1 \right)}^{2}}{{\left( 2n-3 \right)}^{2}}} \\\ \end{aligned}$$ Now let us divide the numerator and denominator inside limit by ${{n}^{4}}$ $==\dfrac{25}{32}{{\lim }_{n\to \infty }}.\dfrac{\dfrac{{{n}^{2}}}{{{n}^{2}}}\left( \dfrac{n}{n}-\dfrac{1}{n} \right)\left( \dfrac{n}{n}-\dfrac{3}{n} \right)}{{{\left( \dfrac{2n}{n}-\dfrac{1}{n} \right)}^{2}}{{\left( \dfrac{2n}{n}-\dfrac{3}{n} \right)}^{2}}}$ $$\begin{aligned} & =\dfrac{25}{32}.\dfrac{(1)(1-0)(1-0)}{{{\left( 2-0 \right)}^{2}}{{\left( 2-0 \right)}^{2}}} \\\ & =\dfrac{25}{32}\dfrac{(1)(1-0)(1-0)}{16} \\\ & =\dfrac{25}{512} \\\ \end{aligned}$$ Hence the required probability is $\dfrac{25}{512}$ **So, the correct answer is “Option A”.** **Note:** While multiplying ${{n}^{4}}$ to numerator and denominator note that we multiply it to the whole numerator and not to each term. Also since in the denominator if we multiply ${{n}^{4}}$ and take it inside bracket of ${{(2n-1)}^{2}}$ note that the power will become ${{n}^{2}}$and hence we will get $\left( \dfrac{2n-1}{{{n}^{2}}} \right)$