Question

An urn contains six blue balls and seven yellow balls. If four balls are selected randomly without being​ replaced, what is the probability that of the balls​ selected, three of them will be blue and one of them will be​ yellow?

Answer 1

Answer: 0.0490

Explanation:

Given : Number of blue balls = 6

Number of yellow balls = 7

Total number of balls = 13

Total number of ways for selecting 4 balls from 13 balls :-

`^(13)P_4=(13!)/((13-4)!)=17160`

The number of ways of selecting 3 blue and one yellow :-

`^(6)P_3*^(7)P_1=(6!)/(3!)*(7!)/(6!)=840`

Now, the probability that of the balls​ selected, three of them will be blue and one of them will be​ yellow :-

`(840)/(17160)\approx0.0490`

Hence, the required probability is 0.0490 .