(a) A manufacturer produces light bulbs which are tested in the following way. A batch is accepted in either of the following cases:
(i) a first sample of 5 bulbs contains no faulty bulbs ; (ii) a first sample of 5 bulbs contains at least one faulty bulb but a second sample of size 5 has no faulty bulb. If 10% of the bulbs are faulty, what is the probability that the batch is accepted?
(b) A bag contains 15 identical marbles of which 3 are black, Keshi picks a marble at random from the bag and replaces it. If this is repeated 10 times; what is the probability that he :
(i) did not pick a black ball? (ii) picked a black ball at most three times?
(i) a first sample of 5 bulbs contains no faulty bulbs ; (ii) a first sample of 5 bulbs contains at least one faulty bulb but a second sample of size 5 has no faulty bulb. If 10% of the bulbs are faulty, what is the probability that the batch is accepted?
(b) A bag contains 15 identical marbles of which 3 are black, Keshi picks a marble at random from the bag and replaces it. If this is repeated 10 times; what is the probability that he :
(i) did not pick a black ball? (ii) picked a black ball at most three times?
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Correct Answer: Option n
Explanation:
(a) 10% of bulbs are faulty, 90% are not faulty.
P(a bulb is flawless) = 90% = 0.9
(i) p(first sample is flawless and accepted) = \((0.9)^{5} = 0.59049 \approxeq 0.590\)
(ii) P(first sample of 5 bulbs contains at least one faulty bulb) = 1 - 0.59049 = 0.40951\)
Probability that the second sample of size 5 contains flawless bulbs = \((0.9)^{5} = 0.59049\)
\(\therefore\) probability that the first batch is accepted = \(0.4095 \times 0.5905 = 0.2418\)
(b) 3 black, 12 non-black. Total = 15.
p(black marble) = \(\frac{3}{15} = p = \frac{1}{5}\)
p(non- black marble) = \(\frac{12}{15} = q = \frac{4}{5}\)
\((p + q)^{10} = p^{10} + 10p^{9} q + 45p^{8} q^{2} + 120p^{7} q^{3} + 210p^{6} q^{4} + 252p^{5} q^{5} + 210p^{4} q^{6} + 120p^{3} q^{7} + 45p^{2} q^{8} + 10p q^{9} + q^{10}\)
(i) p(no black) = \(q^{10} = (\frac{4}{5})^{10} = \frac{1048576}{9765625}\)
(ii) p(black marble at most 3 times) = p(0 black marble) + p(1 black marble) + p(2 black marbles) + p(3 black marbles)
= \(\frac{1048576}{9765625} + 10(\frac{1}{5})(\frac{4}{5})^{9} + 45(\frac{1}{5})^{2} (\frac{4}{5})^{8} + 120(\frac{1}{5})^{3} (\frac{4}{5})^{7}\)
(a) 10% of bulbs are faulty, 90% are not faulty.
P(a bulb is flawless) = 90% = 0.9
(i) p(first sample is flawless and accepted) = \((0.9)^{5} = 0.59049 \approxeq 0.590\)
(ii) P(first sample of 5 bulbs contains at least one faulty bulb) = 1 - 0.59049 = 0.40951\)
Probability that the second sample of size 5 contains flawless bulbs = \((0.9)^{5} = 0.59049\)
\(\therefore\) probability that the first batch is accepted = \(0.4095 \times 0.5905 = 0.2418\)
(b) 3 black, 12 non-black. Total = 15.
p(black marble) = \(\frac{3}{15} = p = \frac{1}{5}\)
p(non- black marble) = \(\frac{12}{15} = q = \frac{4}{5}\)
\((p + q)^{10} = p^{10} + 10p^{9} q + 45p^{8} q^{2} + 120p^{7} q^{3} + 210p^{6} q^{4} + 252p^{5} q^{5} + 210p^{4} q^{6} + 120p^{3} q^{7} + 45p^{2} q^{8} + 10p q^{9} + q^{10}\)
(i) p(no black) = \(q^{10} = (\frac{4}{5})^{10} = \frac{1048576}{9765625}\)
(ii) p(black marble at most 3 times) = p(0 black marble) + p(1 black marble) + p(2 black marbles) + p(3 black marbles)
= \(\frac{1048576}{9765625} + 10(\frac{1}{5})(\frac{4}{5})^{9} + 45(\frac{1}{5})^{2} (\frac{4}{5})^{8} + 120(\frac{1}{5})^{3} (\frac{4}{5})^{7}\)