Question

A bag contains a very large number of white and black marbles in the ratio $1:2$. Two samples of marbles, $5$ each , are picked up. The probability that the first sample contains exactly one black and second one contains exactly three marbles is:
A. $\dfrac{{10}}{{{3^9}}}$
B. $\dfrac{{1600}}{{{3^{10}}}}$
C. $\dfrac{{800}}{{{3^{10}}}}$
D. $\dfrac{{10}}{{{3^5}}}$

Answer 1

Hint: Here, we are given the ratio of white and black marbles in a bag. Two samples of five marbles each are picked up and we have to find the probability that the first one contains exactly one lack and the second one contains exactly three black marbles. So, we first find the probability of picking one black marble from the bag. Then, we make use of the conditional probability concept to determine the probability of the simultaneous events.

Complete step by step answer:
In the given question, white and black marbles are present in the ratio $1:2$. Let there be a total of $3x$ marbles in the bag. So, there will be x white and $2x$ black marbles.
Hence, the probability of picking up one black marble from the bag $ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Number of total outcomes}}}}=\dfrac{{\text{x}}}{{{\text{3x}}}} = \dfrac{1}{3}$
Now, the number of marbles in the first sample is five and it consists of exactly one black marble. So, there are a total of five options for a black ball. So, probability of exactly one black marble in the first sample is
$\left( {^5{C_1}} \right)\left( {\dfrac{2}{3}} \right){\left( {\dfrac{1}{3}} \right)^4}$

Also, the number of marbles in the first sample is five and it consists of exactly three black marbles. So, there are a total of five balls out of which three are black in colour. So, probability of exactly three black marbles in the second sample is,
$\left( {^5{C_3}} \right){\left( {\dfrac{2}{3}} \right)^3}{\left( {\dfrac{1}{3}} \right)^2}$

Now, for both the events to happen, the probability is the product of their individual probabilities as they are independent events.
So, we get the probability as $ = \left( {^5{C_1}} \right)\left( {\dfrac{2}{3}} \right){\left( {\dfrac{1}{3}} \right)^4} \times \left( {^5{C_3}} \right){\left( {\dfrac{2}{3}} \right)^3}{\left( {\dfrac{1}{3}} \right)^2}$
Substituting the values and simplifying the calculations, we get,
$\left( 5 \right)\left( {\dfrac{2}{3}} \right)\left( {\dfrac{1}{{81}}} \right) \times \left( {10} \right)\left( {\dfrac{8}{{27}}} \right)\left( {\dfrac{1}{9}} \right)$
$\Rightarrow \dfrac{{10}}{{243}} \times \dfrac{{80}}{{243}}$
$\therefore \dfrac{{800}}{{{3^{10}}}}$

So, option C is the correct answer.

Note: These problems are the combinations of ratio and probability, so, the concepts of both of the topics are used in these. We must know the concept of probability of two independent events is the product of the probabilities of the individual events. We must take care of the calculations in order to match our answer in the options. We should know the combination formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ to simplify the calculations and get to the final answer.