?
A bag contains 6 red balls 11 yellow balls and 5 pink balls. If two balls are drawn at random from the bag. One after another what is the probability that the first ball is red and second ball is yellow?
Answer & Solution
Correct Answer:
Option
C
Number of red balls = 6
Number of yellow balls = 11
Number of pink balls = 5
Total number of balls = 6 + 11 + 5 = 22
Total possible outcomes
$$n(E) = {}^{22}\mathop C\nolimits_2 = \frac{{22!}}{{2!(22 - 2)!}}$$
$$ = \frac{{22!}}{{2! \times 20!}} $$
$$ = \frac{{22 \times 21}}{{2 \times 1}} $$
$$=$$ 231
Number of favourable outcomes
$$n(S) = {}^6\mathop C\nolimits_1 \times {}^{11}\mathop C\nolimits_1 $$ = 6 × 11 = 66
Required probability = $$\frac{{n(E)}}{{n(S)}}$$ $$ = \frac{{66}}{{231}}$$ $$ = \frac{1}{7}$$
Number of yellow balls = 11
Number of pink balls = 5
Total number of balls = 6 + 11 + 5 = 22
Total possible outcomes
$$n(E) = {}^{22}\mathop C\nolimits_2 = \frac{{22!}}{{2!(22 - 2)!}}$$
$$ = \frac{{22!}}{{2! \times 20!}} $$
$$ = \frac{{22 \times 21}}{{2 \times 1}} $$
$$=$$ 231
Number of favourable outcomes
$$n(S) = {}^6\mathop C\nolimits_1 \times {}^{11}\mathop C\nolimits_1 $$ = 6 × 11 = 66
Required probability = $$\frac{{n(E)}}{{n(S)}}$$ $$ = \frac{{66}}{{231}}$$ $$ = \frac{1}{7}$$
Join the Discussion
Login to post a comment or share your explanation.
Login6/22 × 11/21 = 1/7
as the order of picking of balls are give first is red and second is yellow instead of 22c2
we should select 22p2 as sequence of selection is given
required probability is [(6c1)*(11c1)]/(22p2)=1/7