Question
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: Option B
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 = $$\frac{{n(E)}}{{n(S)}}$$ $$ = \frac{{66}}{{231}}$$ $$ = \frac{2}{7}$$
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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 = $$\frac{{n(E)}}{{n(S)}}$$ $$ = \frac{{66}}{{231}}$$ $$ = \frac{2}{7}$$
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