91Ó°ÊÓ

A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable batteries have been found. Out of all batteries tested, 90% of them have acceptable voltage. The random variable\(Y\)denote the number of batteries that must be tested.

a.

The probability that a battery has acceptable voltage is \(P\left( A \right) = 0.9\).

So, the probability of a battery has not acceptable voltage is,

\(\begin{aligned}P\left( U \right) &= 1 - P\left( A \right)\\ &= 1 - 0.9\\ &= 0.1\end{aligned}\)

The experiment is performed until two acceptable batteries have to be found. Since the random variable denote the number of batteries that must be tested, so \(Y = 2\) means both selected batteries have accepted voltage i.e.\(\left\{ {AA} \right\}\).

As both batteries are chosen independently, the required probability is computed as:

\(\begin{aligned}P\left( {Y = 2} \right) &= P\left( {AA} \right)\\ &= P\left( A \right) \times P\left( A \right)\\ &= 0.9 \times 0.9\\ &= 0.81\end{aligned}\)

Thus, the required probability is 0.81.

b.

In this case, when\(Y = 3\) it means the number of trials is 3 andthe last battery must have accepted voltage. The possible pairs are\(\left\{ {\left( {UAA} \right),\left( {AUA} \right)} \right\}\).

As we do not consider the pair\(\left( {AAU} \right)\)because the last trail must be of acceptable voltage.

So, the probability is computed as,

\(\begin{aligned}P\left( {Y = 3} \right) &= P\left( {UAA} \right) + P\left( {AUA} \right)\\ &= \left( {\left( {0.1} \right){{\left( {0.9} \right)}^2} + \left( {0.1} \right){{\left( {0.9} \right)}^2}} \right)\\ &= 0.081 + 0.081\\ &= 0.1620\end{aligned}\)

Thus, the required probability is 0.1620.

c.

In this case, when\(Y = 5\) it means the number of trials is 5 and the last battery must have accepted voltage.

There are only four possible pairs as below:

\(\left\{ {\left( {UUUAA} \right),\left( {UUAUA} \right),\left( {UAUUA} \right),\left( {AUUUA} \right)} \right\}\)

So, the required probability i.e.\(P\left( {Y = 5} \right)\) is computed as,

\(\begin{aligned}P\left( {Y = 5} \right) &= P\left( {UUUAA} \right) + P\left( {UUAUA} \right) + P\left( {UAUUA} \right) + P\left( {AUUUA} \right)\\ &= \left( {{{\left( {0.1} \right)}^3}{{\left( {0.9} \right)}^2} + {{\left( {0.1} \right)}^3}{{\left( {0.9} \right)}^2} + {{\left( {0.1} \right)}^3}{{\left( {0.9} \right)}^2} + {{\left( {0.1} \right)}^3}{{\left( {0.9} \right)}^2}} \right)\\ &= \left( {0.0008 + 0.0008 + 0.0008 + 0.0008} \right)\\ &= 0.0032\end{aligned}\)

Thus, the required probability is 0.0032.

d.

For part(a) of the problem, the minimum accepted batteries are two or two trials are conducted.

Since the experiment is being conducted until two acceptable ones have been found. So, the number of success is 2 and number of failures is 1.

It can be rewritten as:

\(P\left( {AA} \right) = \left( {2 - 1} \right){\left( {P\left( A \right)} \right)^2}{\left( {P\left( U \right)} \right)^{2 - 2}}\)

For part (b) of the problem, an experiment of three trials is being conducted until two acceptable voltages have been found or\(Y = 3\).

The number of success is 2 and number of failures is 1.

It can be rewritten as:

\(P\left( {UAA} \right) + P\left( {AUA} \right) = \left( {3 - 1} \right){\left( {P\left( A \right)} \right)^2}{\left( {P\left( U \right)} \right)^{3 - 2}}\)

For part (c) of the problem, an experiment of five trials is being conducted until two acceptable voltages have been found or\(Y = 5\).

The number of success is 2 and number of failures is 3.

It can be rewritten as:

\(P\left( {UUUAA} \right) + P\left( {UUAUA} \right) + P\left( {UAUUA} \right) + P\left( {AUUUA} \right) = \left( {5 - 1} \right){\left( {P\left( A \right)} \right)^2}{\left( {P\left( U \right)} \right)^{5 - 2}}\)

Therefore, the general formulaof\(P\left( Y \right)\)can be written as:

\(P\left( Y \right) = \left( {Y - 1} \right)P{\left( A \right)^2}P{\left( U \right)^{Y - 2}}\)