Q. 6.6 - A First Course in Probability

Question

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defective ones are identified. Denote by N₁ the number of tests made until the first defective is identified and by N₂ the number of additional tests until the second defective is identified. Find the joint probability mass function of N₁ and N₂. 

Step-by-Step Solution

Answer

The probability is $P (N_1=k, N_2=l)= P(N_2=l, N_1=k), P(N_1=k)=\frac{1}{\left({\displaystyle \frac{5-k}{1}}\right)}\times \frac{{\displaystyle \frac{5-k}{1}}}{{\displaystyle \frac{5}{2}}}=\frac{1}{{\displaystyle \frac{5}{2}}}$

Step 1: Content Introduction

Probability implies possibility A part of science manages the event of an irregular occasion. The worth is communicated from zero to one.

Step 2: Content Explanation

Firstly, observe the random variables N₁ assume values in (1,....,4). The distribution of N₁ is simple. Take any $k\in \{1,....,4\}$. In total there exist $\frac{5}{2}$, to permute transistors in an order of detecting.  We want that first defective transistors show on kth place. It mean that on first k-1 place have to be non-defective transistors. On remaining places arbitrarily put on remaining defective transistors. Hence,

$P (N_1=k)=\frac{{\displaystyle \left(\frac{5-k}{1}\right)}}{{\displaystyle \left(\frac{5}{2}\right)}}$

Step 3: Conclusion

The first distribution of N₂ is a bit confusing. Instead, we are going to find the conditional distribution of N₂ given that N₁=k. In this case we have, $N_2\in \{1,....,5-k\}$. There exist $\left(\frac{5-k}{1}\right)$ ways to arrange remaining transistors. If we want that we will need L additional steps to obtain the second defective transistor, there is only one possible arrangement for that option.

Hence, $P(N_2=l, N_1=k)=\frac{1}{\left({\displaystyle \frac{5-k}{1}}\right)}$