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?
A. $$\frac{1}{14}$$
B. $$\frac{2}{7}$$
C. $$\frac{1}{7}$$
D. $$\frac{3}{14}$$
E. None of these
Answer: Option C
Solution(By Examveda Team)
Number of red balls = 6Number 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 = $$\frac{{n(E)}}{{n(S)}}$$ $$ = \frac{{66}}{{231}}$$ $$ = \frac{1}{7}$$
This Question Belongs to Arithmetic Ability >> Probability
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Comments ( 10 )
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Answer should be 1/7.
Your answer is wrong. The correct answer will be 1/7.
6/22 × 11/21 = 1/7
It should be the correct answer 1/7 because of clearly mensioned 1st is red and 2nd is yellow, not replacement with succession.
it should be 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
1/7
if without replacement case is there then its answer should be 1/7
It should be 1/7 as mentioned 1st is red and 2nd is yellow
6/22 X 11/21 = 1/7
Result will be 1/7
6/22 is the probability of getting 1red ball and 11/21is the probability of getting yellow ball in second time.and the answer to this question is multiplication of these two outcomes