# Question 2 Exercise 6.4

Solutions of Question 2 of Exercise 6.4 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

A bag contain $4$ white, $5$ red and $6$ green balls. $3$ balls are drawn at random. What is the probability that all are green?

Total number of balls are: $4+5+6=15$ balls

Total number of ways drawing three balls at random are: $${ }^{15} C_3=\dfrac{15 !}{(15-3) ! 3 !}=455$$

All are green

The total number ways of favorable outcomes for green balls are: $${ }^6 C_4=\dfrac{6 !}{(6-4) ! 4 !}=15$$

Now the probability that all balls are green is: $$=\dfrac{15}{455}=\dfrac{3}{91}$$

A bag contain $4$ white, $5$ red and $6$ green balls. $3$ balls are drawn at random. What is the probability that all are white?

Total number of balls are: $4+5+6=15$ balls

Total number of ways drawing three balls at random are: $${ }^{15} C_3=\dfrac{15 !}{(15-3) ! 3 !}=455$$

All are white

The total number of ways drawing three white balls are: $${ }^4 C_3=\dfrac{4 !}{(4-3) ! 3 !}=4$$ Thus the probability that all balls are white is: $=\dfrac{4}{455}$.