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A box has 20 pens of which 2 are defective. Calculate the probabilitythat out of 5 pens drawn one by one with replacement, at most 2 aredefective. |
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Answer» Probability of selecting 2 defective pens=2/10=1/10 Probability of selecting pens which are not defective=18/20=9/10 1)Number of defective pen=0 `P(A)=5C_0(1/10)^0(9/10)^5` 2)Number of defective pen=1 `P(B)=5C_1(1/10)^1(9/10)^4` 3)Number of defective pen=2 `P(C)=5C_2(1/10)^2(9/10)^3` Total probability that at most 2 defective pens `=P(A)+P(B)+P(C)` `5C_0(1/10)^0(9/10)^5+5C_1(1/10)^1(9/10)^4+5C_2(1/10)^2(9/10)^3` `=(9/10)^3*34/25`. |
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