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The probability that a certain kind of component will survive a check test is 0.5. Find the probability that exactly two of the next four components tested will survive. |
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Answer» Let X = number of tested components survive. p = probability that the component survive the check test. `therefore" "p=0.5 =(5)/(10)=(1)/(2)` `therefore" "q=1-p=1-(1)/(2)=(1)/(2)` Given n = 4 `therefore X ~ B(4, 1//4)` The probability mass function as : `P(X=x)=.^(n)C_(x)p^(x).q^(n-x)` i.e., `p(x)=.^(4)C_(x)((1)/(2))^(2x)((1)/(2))^(4-x)` P(exactly 2 components survive) `=P(X=2)=p(2)` `=.^(4)C_(2)((1)/(2))^(2)((1)/(2))^(4-2)` `=(4!)/(2!.2!).((1)/(2))^(2).((1)/(2))^(2)` `=(4.3.2!)/(2.1.2!)xx((1)/(2))^(4)` `=(6)/(16)=0.375` Hence, the probabillity that exactly 2 of the 4 tested components survive is 0.375. |
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