Question 1150083
Part (A):
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Probability of 2 defective bulbs in the sample: {{{(10/30) * (9/29)}}} = 9/87...(can be reduced to 3/29)
Probability of 1 defective bulb in the sample: {{{(10/30) * (20/29) * (2!/(1!*1!))}}} = 40/87
Probability of 0 defective bulbs in the sample: {{{(20/30) * (19/29)}}} = 38/87
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Part (B):
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2 * 9/87 = 18/87
1 * 40/87 = 40/87
0 * 38/87 = 0
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18/87 + 40/87 + 0 = 58/87 = 2/3
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The number of expected defective bulbs in the sample is 2/3...or, 0.6667.