Question

In a box, there are eight yellow and four black balls. If three balls are drawn at random, what is the probability that two are yellow and one black?

Answer 1

hope this helps you ☺️☺️

Answer 2

Answer: The answer is $\dfrac{28}{55}.$

Step-by-step explanation: There are total 12 balls in the box, out of which 8 are yellow in colour and 4 are black in colour.

And, we are to draw 3 balls at random from the bag. Then, the number of ways in which we can do so with 2 balls yellow and one black is given by

$m=\dfrac{8!}{2!(8-2)!}\times \dfrac{4!}{1!(4-1)!}=\dfrac{8!}{2!6!}\times \dfrac{4!}{1!3!}=\dfrac{8\times 7}{2\times 1}\times 4=112.$

Also, the total number of ways is given by

$n=\dfrac{12!}{3!(12-3)!}=\dfrac{12!}{3!9!}=\dfrac{12\times 11\times 10}{3\times 2\times 1}=220.$

Therefore, the required probability is

$p=\dfrac{no.~of~favourible~ways}{total~no.~of~ways}=\dfrac{112}{220}=\dfrac{28}{55}.$

Thus, the required probability is $\dfrac{28}{55}.$