(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...
Question 1 Report
(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?
Answer Details
(a) With \(10\%\) faulty, \(P(\text{a bulb is good}) = 0.9\). For a sample of 5, \(P(\text{no faulty}) = (0.9)^5 = 0.59049\).
The batch is accepted if the first sample has no faulty, OR the first sample has at least one faulty but the second sample has none:
\[P(\text{accept}) = (0.9)^5 + \big[1 - (0.9)^5\big](0.9)^5 = 0.59049 + (0.40951)(0.59049).\]
\[= 0.59049 + 0.241813 = 0.832303 \approx 0.832.\]
(b) Bag of 15 marbles, 3 black, so \(P(\text{black}) = \tfrac{3}{15} = 0.2\), \(P(\text{not black}) = 0.8\), with replacement, \(n = 10\).
(i) Did not pick a black at all: \((0.8)^{10} = 0.1073742 \approx 0.107\).
(ii) Picked black at most three times \(= P(0)+P(1)+P(2)+P(3)\):
\[P(0)=0.107374,\ P(1)=\binom{10}{1}(0.2)(0.8)^9=0.268435,\]
\[P(2)=\binom{10}{2}(0.2)^2(0.8)^8=0.301990,\ P(3)=\binom{10}{3}(0.2)^3(0.8)^7=0.201327.\]
Sum \(= 0.879\) (to three decimal places).
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