91Ó°ÊÓ

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

Firstly, observe the random variables N₁assume values in (1,....,4). The distribution of N₁ is simple. Take any $k∈\{1,....,4\}$. 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,

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

The first distribution of N₂is a bit confusing. Instead, we are going to find the conditional distribution of N₂given that N₁=k. In this case we have, $N_2∈\{1,....,5-k\}$. 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,$P(N_2=l,N_1=k)=\frac{1}{\left({\displaystyle \frac{5-k}{1}}\right)}$