Question

A purchaser of transistors buys them in lots of $20$. It is his policy to randomly inspect $4$ components from a lot and to accept the lot only if all $4$ are nondefective. If each component in a lot is, independently, defective with probability $.1,$ what proportion of lots is rejected?

Short answer

$P($ Reject lot$)=34.39\%$

Step-by-step solution

Step 1:Given information

Given in the question that a purchaser of transistors buys them in lots of 20. It is his policy to randomly inspect 4 components from a lot and to accept the lot only if all 4 are nondefective.

Step 2:Explanation

Given:

$n=$number of trials$=4$

$p=$probability of success$=0.1$

The number of successes among a fixed number of independent trials with a constant probability of success follows a binomial distribution.

Definition binomial probability:

$P(X=k)=\left(\begin{array}{l}n\\ k\end{array}\right)·p^k·(1-p{)}^{n-k}$

The lot is accepted if all $4$selected components are not defective. Evaluate the definition of binomial probability at $k=0$

$P($Accept lot $)=P(X=0)$

$=\left(\begin{array}{l}4\\ 0\end{array}\right)·0.1^0·(1-0.44{)}^{4-0}$

$=\frac{4!}{0!(4-0)!}·0.1^0·0.{56}^4$

$=1·1·0.{56}^4$

$\approx 0.6561$

Step 3:Complement Rule

Use the Complement rule:

$P\left(A^c\right)=P($not $A)=1-P(A)$

$P($Reject lot $)=1-P($Accept lot $)$

$=1-0.6561$

$=0.3439$

$=34.39\%$

Final answer

$P($Reject lot$)=$$34.39\%$