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.
Question 2(i)
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?
Solution
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: $$=\dfrac{15}{455}=\dfrac{3}{91}$$
Question 2(ii)
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?
Solution
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: $=\dfrac{4}{455}$.
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