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Brookes · Answer 1 · 2023-04-18T09:51:01+0000

Given 16 transistors, 3 of which are defective, 13 are not defective

Part A: All are defective.

With replacement

$\begin{gathered} P=(3)/(16)\cdot(3)/(16)\cdot(3)/(16) \\ P=(27)/(4096) \\ \\ \text{The probability that all are defective with replacement is }(27)/(4096) \end{gathered}$

Without replacement

$\begin{gathered} P=(3)/(16)\cdot(2)/(15)\cdot(1)/(14) \\ P=(143)/(280) \\ \\ \text{The probability that all are defective without replacement is }(1)/(560) \end{gathered}$

Part B: None are defective

With replacement

$\begin{gathered} P=(13)/(16)\cdot(13)/(16)\cdot(13)/(16) \\ P=(2197)/(4096) \\ \\ \text{Therefore, the probability that none are defective with replacement is }(2197)/(4096) \end{gathered}$

Without replacement

$\begin{gathered} P=(13)/(16)\cdot(12)/(15)\cdot(11)/(14) \\ P=(143)/(280) \\ \\ \text{Therefore, the probability that none are defective without replacement is }(143)/(280) \end{gathered}$

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