Question

A bag contains 5 white and 7 black balls, if two balls are drawn what is the chance that one is white and the other is black?

Answer 1

Hint: In order to solve this question, we should know the most basic formula of probability that is, $\text{Probability = }\dfrac{\text{Favorable Number of Outcomes}}{\text{Total Number of Outcomes}}$. Also, we have to keep in mind that both the balls are drawn at the same time. So, the total number of outcomes and favorable outcomes will be calculated using the combination.

Complete step-by-step answer:
In this question, we have to find the probability of drawing two balls such that one is white and the other is black. To solve this question, we need to know the concept of probability, which is the ratio of favorable outcomes to the total number of outcomes. Mathematically, we can write it as,
$\text{Probability = }\dfrac{\text{Favorable Number of Outcomes}}{\text{Total Number of Outcomes}}$
Now, we have been given that two balls are drawn, so to find the total number of ways of selecting two balls from 12 balls, we will use the combination formula which is used to select the number of ways of combining the things irrespective of their orders. So, the number of ways of choosing 2 balls out of 5 white and 7 black balls, that is 2 out of 12 balls is $^{12}{{C}_{2}}$.
Total Number of outcomes = $^{12}{{C}_{2}}$
Now, we will find the number of ways of choosing one white and one black ball. We know that there are $^{5}{{C}_{1}}$ and $^{7}{{C}_{1}}$ ways of choosing one white and one black ball respectively. So, we can say,
Favorable Number of Outcomes $={{\text{ }}^{5}}{{C}_{1}}{{\times }^{7}}{{C}_{1}}$
Now, we will put the values of the total number of outcomes and a favorable number of outcomes in the formula of probability. Therefore, we will get,
$\text{Probability}=\dfrac{^{5}{{C}_{1}}\times {{\text{ }}^{7}}{{C}_{1}}}{^{12}{{C}_{2}}}$
Now, we know that $^{n}{{C}_{r}}$ can be expanded as $\dfrac{n!}{r!\left( n-r \right)!}$. So, we can write the probability as
$\text{Probability}=\dfrac{\dfrac{5!}{1!\left( 5-1 \right)!}.\dfrac{7!}{1!\left( 7-1 \right)!}}{\dfrac{12!}{2!\left( 12-2 \right)!}}$
$\text{Probability}=\dfrac{\dfrac{5!}{1!4!}.\dfrac{7!}{1!6!}}{\dfrac{12!}{2!10!}}$
And we can further write it as
$\text{Probability}=\dfrac{5!\text{ }7!\text{ }2!\text{ }10!}{1!\text{ }4!\text{ }1!\text{ }6!\text{ }12!}$
Now, we know that,
$n!=\left( n \right)\left( n-1 \right)\left( n-2 \right)....\left( 1 \right)$
So, we can write,
$\text{Probability}=\dfrac{5\times 4!\times 7\times 6!\times 2!\times 10!}{1!\times 4!\times 1!\times 6!\times 12\times 11\times 10!}$
$\text{Probability}=\dfrac{5\times 7\times 2}{12\times 11}$
$\text{Probability}=\dfrac{35}{66}$
Hence, we can say that the probability of drawing one white and one black ball is $\dfrac{35}{66}$.

Note: In this question, there are high possibilities that we might make a mistake by assuming that the balls are drawn out one by one by replacing the wrong one. Balls are not replaced in the box. So, the total number of outcomes will be reduced by 1 after one ball is drawn out.