91Ó°ÊÓ

Two circuit boards are selected from each five batches for inspection of incoming materials for defects.

a.

There is selection of two boards from each batch having five boards in order to find the possible outcomes.

As order does not matter, use combination formula to find the number of ways that two boards are drawn from each batch of 5 boards.

\(^n{C_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}}\)

Here, the total number of objects are,\(n = 5\)and the number of choosing objects,\(r = 2\).

\(\begin{aligned}^5{C_2} &= \frac{{5!}}{{2!\left( {5 - 2} \right)!}}\\ &= \frac{{5 \times 4 \times 3!}}{{\left( {2!} \right)\left( {3!} \right)}}\\ &= 10\end{aligned}\)

The 10 possible outcomes of the selection process are described by pairs as follows:

\(\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {4,5} \right)\)

Here, the pair (1,2) represent the selection of boards 1 and 2.

b.

Out of 5 boards in a batch, only board 1 and 2 are defective so to compute the probability\(P\left( {X = 0} \right)\), avoid the pairs that consist of board 1 and board 2 from 10 possible pairs.

The remaining pairs are\(\left\{ {\left( {3,4} \right),\left( {3,5} \right),\left( {4,5} \right)} \right\}\).

So, the probability is computed as,

\(\begin{aligned}P\left( {X = 0} \right) &= \frac{3}{{10}}\\ &= 0.30\end{aligned\)

Similarly, compute the probability\(P\left( {X = 1} \right)\). In this case, either one of the pair is board 1 or board 2.

The possible pairs are\(\left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right)} \right\}\).

So, the probability is computed as,

\(\begin{array}{c}P\left( {X = 1} \right) = \frac{6}{{10}}\\ = 0.60\end{array}\)

Similarly, compute the probability\(P\left( {X = 2} \right)\). In this case, a pair consist of both board 1 and board 2.

The possible pair is\(\left\{ {\left( {1,2} \right)} \right\}\).

So, the probability is computed as,

\(\begin{aligned}P\left( {X = 2} \right) & = \frac{1}{{10}}\\ &= 0.10\end{aligned}\)

Therefore, the probability distribution of X is given as,

c.

From the problem, the cdf\(F\left( {X = x} \right)\) can be defined as,

\(F\left( {X = x} \right) = P\left( {X \le x} \right)\)

First, compute\(F\left( 0 \right)\) using the above formula:

\(\begin{aligned}F\left( {X = 0} \right) &= P\left( {X \le 0} \right)\\ &= P\left( {X = 0} \right)\\ = 0.30\end{aligned}\)

Then, compute\(F\left( 1 \right)\) using the above formula:

\(\begin{aligned}F\left( {X = 1} \right) &= P\left( {X \le 1} \right)\\ &= P\left( {X = 0} \right) + P\left( {X = 1} \right)\\ &= 0.30 + 0.60\\ &= 0.90\end{aligned}\)

Finally compute\(F\left( 2 \right)\) using the above formula:

\(\begin{aligned}F\left( {X = 2} \right) &= P\left( {X \le 2} \right)\\ &= P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right)\\ &= 0.30 + 0.60 + 0.10\\ &= 1\end{aligned}\)

Therefore, the cumulative distribution functionof X is given as,

\(F\left( x \right) = \left\{ \begin{array}{l}0,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,x < 0\\0.3\,\,\,\,\,\,\,\,{\rm{when}}\,0 \le x < 1\\0.9\,\,\,\,\,\,\,\,{\rm{when}}\,1 \le x \le 2\\0.1\,\,\,\,\,\,\,\,\,{\rm{when}}\,x \ge 2\end{array} \right.\)