====== 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: $=\dfrac{4}{455}$.



====Go To====
<text align="left"><btn type="primary">[[math-11-kpk:sol:unit06:ex6-4-p1 |< Question 1 ]]</btn></text>
<text align="right"><btn type="success">[[math-11-kpk:sol:unit06:ex6-4-p3|Question 3 >]]</btn></text>