Question

A card is drawn from a pack. The card is replaced, and the pack is reshuffled. If this is done six times, the probability that $2$ hearts, $2$ diamonds, and $2$ black cards are drawn is
1. $90.{\left( {\frac{1}{2}} \right)^4}$
2. $\left( {\frac{{45}}{2}} \right){\left( {\frac{3}{4}} \right)^4}$
3. $\frac{{90}}{{{2^{10}}}}$
4. None of these

Answer 1

Hint: In this question, we are given that a card is drawn from the pack then replaced the card and reshuffled in the pack. This process was repeated six times. We have to find the probability that $2$ hearts, $2$ diamonds, and $2$ black cards are drawn. First, find the probability of one card being drawn off each type of card using the probability formula. Then, using combination find the number of ways the card is drawn. In the last multiply all the ways and the square of each probability to calculate the required probability.

Formula Used:
Probability formula – The probability of an event occurring is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
\[Probability = \frac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
Combination formula –
${}^n{C_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}},n! = n\left( {n - 1} \right)\left( {n - 2} \right).......$

Complete step by step Solution:
Total number of cards in a pack $ = 52$
Total number of heart cards $ = 13$
Total number of diamond cards $ = 13$
Total number of black cards $ = 26$
Using the probability formula,
\[Probability = \frac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
Probability of getting a heart, diamond, and black card in one draw are $\frac{{13}}{{52}},\frac{{13}}{{52}},\frac{{26}}{{52}}$ respectively.
Now, the card is replaced, and the pack is reshuffled happened six times,
It implies that, number of ways of the cards drawn are $2$ hearts, $2$ diamonds and $2$ black cards are ${}^6{C_2},{}^4{C_2},{}^2{C_2}$respectively.
The required probability will be ${}^6{C_2} \times {\left( {\frac{{13}}{{52}}} \right)^2} \times {}^4{C_2} \times {\left( {\frac{{13}}{{52}}} \right)^2} \times {}^2{C_2} \times {\left( {\frac{{26}}{{52}}} \right)^2}$
\[ = {}^6{C_2} \times {\left( {\frac{1}{4}} \right)^2} \times {}^4{C_2} \times {\left( {\frac{1}{4}} \right)^2} \times {}^2{C_2} \times {\left( {\frac{1}{2}} \right)^2}\]
\[ = 90.{\left( {\frac{1}{2}} \right)^{10}}\]
\[ = \frac{{90}}{{{2^{10}}}}\]


Hence, the correct option is 3.

Note: The key concept involved in solving this problem is a good knowledge of combination. Students must know that combinations are a method of calculating the total outcomes of an event where the order of the outcomes is irrelevant.