Out of the 24 apples in a box, 6 are bad. If three apples are taken from the box at random, with replacement, find the probability that :
(a) the first two are good and the third is bad ;
(b) all three are bad ;
(c) all the three are good.
(a) the first two are good and the third is bad ;
(b) all three are bad ;
(c) all the three are good.
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Correct Answer: Option n
Explanation:
(a) \(P(Good) = \frac{18}{24} = \frac{3}{4}\)
\(P(Bad) = \frac{6}{24} = \frac{1}{4}\)
P(2 are good and 1 bad) = \(\frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}\)
= \(\frac{9}{64}\)
(b) P(the 3 are bad) = \(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}\)
= \(\frac{1}{64}\)
(c) P(the 3 are good) = \(\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}\)
= \(\frac{27}{64}\)
(a) \(P(Good) = \frac{18}{24} = \frac{3}{4}\)
\(P(Bad) = \frac{6}{24} = \frac{1}{4}\)
P(2 are good and 1 bad) = \(\frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}\)
= \(\frac{9}{64}\)
(b) P(the 3 are bad) = \(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}\)
= \(\frac{1}{64}\)
(c) P(the 3 are good) = \(\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}\)
= \(\frac{27}{64}\)