Question

A group of four components is known to contain two defectives. An inspector tests the components one at a time until the two defectives are located. Let X denote the number of the test on which the second defective is found. Find the probability distribution for X.

Answer 1

Solution

Find the probability distribution for X.

X=The number of the test on which the second defective is found.

\( \bullet \hspace{2mm} \)The possibilities of X are : \( \hspace{2mm}X=2,3,4,.. \)

\( \oplus\hspace{2mm} X=2 : \Big\{(DD)\Big\} \) D=Defictive ,  N=Not defective

\( \implies P(X=2)=\frac{2}{4}\times \frac{1}{3}=\frac{1}{6} \)

\( \oplus\hspace{2mm} X=3 : \Big\{(DND),(NDD)\Big\} \)

\( \implies P(X=3)=2\times \frac{2}{4}\times \frac{2}{3}\times \frac{1}{2}=\frac{1}{3} \)

\( \oplus\hspace{2mm} X=4 : \Big\{(DNND),(NNDD),(NDND)\Big\} \)

\( \implies P(X=4)=3\times \frac{2}{4}\times \frac{2}{3}\times \frac{1}{2}=\frac{1}{2} \)

Therefore.