Question

A bag consists of 7 red balls, 9 white balls and 8 green balls. Three balls are taken out of the bag at random one-by-one. What is the probability that the third ball is neither green nor white if the first two balls are red?

• A. $\frac{2}{23}$
• B. $\frac{5}{22}$
• C. $\frac{7}{25}$
• D. $1$

Answer 1

The correct option is B $\frac{5}{22}$

Given, Number of red balls in the bag = 7
Number of white balls in the bag = 9
And, Number of green balls in the bag = 8

Total number of balls in the bag = 7 + 9 + 8 = 24

Now, Three balls are taken out randomly one by one. The first two balls are red. So, for the last draw, the number of red balls left = 5.
And, total number of balls left for the last draw = 22

The last ball should neither be white nor green.
Or, it must be red.
So, the probability that the last ball drawn is red
$=\frac{Totalnumberofredballsremaining}{Totalnumberofballsremaining}=\frac{5}{22}$