A box contains 16 bulbs out of which 4 bulbs are defective. 3 bulbs are drawn one by one with replacement. Find the mean and variance of the probability distribution of the number of defective bulbs drawn.
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Three bulbs drawn one by one without replacement is the same as drawing 3 bulbs simultaneously. Let X = number of defective bulbs in a lot of 3 bulbs drawn. Then, X=0,1,2 or 3. =12C316C3=(12×11×103×2×1×3×2×116×15×14)=1128.
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