Answer to Question #231099 in Statistics and Probability for syed

Question

A box contains 10 transitors of which 2 are defective . A transistor is selected at

random from the box and tested until a nondefective one is chosen. Then the

expected number of transistors to be chosen is

Answer 1

We will consider a more general situation, when the box contains $n$ transitors, of which $m$ are non-defective. The first non-defective transistor can be found in the steps $1, 2, ... , n-m$ or $n-m+1$ .

For all $k=1,\dots,n-m+1$, let $p_k$ denotes the probability that the first non-defective transistor will be found in the k-th step. We can think that all the transistors are numbered with numbers 1, 2, ..., n, and the number of each transistor is the same as the number of the step in which it will be chosen.

The probability that a non-defective transistor will be found at the k-th step equals to [The number of ways to choose $m-1$ numbers from the set $\{k+1,k+2,\dots,n\}$ for non-defective transistors, starting from the second one]/[The number of ways to choose $m$ numbers from the set $\{1,2,\dots,n\}$ for all non-defective transistors], that is,

(1) $p_k=\binom{n-k}{m-1}/\binom{n}{m}$.

In particular, from this formula it follows that

(2) $\sum\limits_{k=1}^{n-m+1}\binom{n-k}{m-1}=\sum\limits_{k=1}^{n-m+1}p_k\binom{n}{m}=\binom{n}{m}$

By increasing $n$ and $m$ by 1, we also obtain that

(3) $\sum\limits_{k=1}^{n-m+1}\binom{n+1-k}{m}=\binom{n+1}{m+1}$

The expected number of transistors to be chosen is equal to

(4) $\mu=\sum\limits_{k=1}^{n-m+1}kp_k = \sum\limits_{k=1}^{n-m+1}(n+1)p_k- \sum\limits_{k=1}^{n-m+1}(n+1-k)p_k$

The first sum on the right hand side of eq.(4) is equal to

(5) $\sum\limits_{k=1}^{n-m+1}(n+1)p_k= (n+1)\sum\limits_{k=1}^{n-m+1}p_k=n+1$

Since

(6) $(n+1-k)p_k=(n+1-k)\binom{n-k}{m-1}/\binom{n}{m}=$

$=(n+1-k)\frac{(n-k)!}{(m-1)!(n-m-k+1)!}/\binom{n}{m}=$

$=m\frac{(n-k+1)!}{m!(n-m-k+1)!}/\binom{n}{m}=m\binom{n-k+1}{m}/\binom{n}{m}$,

then the second sum on the right hand side of eq.(4) is equal to

(7) $\sum\limits_{k=1}^{n-m+1}(n+1-k)p_k= \sum\limits_{k=1}^{n-m+1}m\binom{n-k+1}{m}/\binom{n}{m}=$

$=\sum\limits_{k=1}^{n-m+1}\binom{n-k+1}{m}\cdot m/\binom{n}{m}=m\binom{n+1}{m+1}/\binom{n}{m}=(n+1)\frac{m}{m+1}$

(eq.(3) was applied).

Finally, by substituting eq.(5) and (7) into eq.(4), we obtain

(8) $\mu=(n+1)-(n+1)\frac{m}{m+1}=\frac{n+1}{m+1}$

This is the answer to the question in the general case. If $n=10$, $m=8$, then $\mu=11/9$.

Answer. $\mu=11/9$