Chapter 9.7, Problem 59E

a.

To determine

To find: the probability that both systems function satisfactorily.

Answer to Problem 59E

  $0.9702$

Explanation of Solution

Given:

The probability that either system will function satisfactorily during a flight is 0.985.

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let E be the event that both systems function satisfactorily.

Here, backup system is independent and so the working of one system doesn’t influence that of the other system.

Also, probability that either system will function satisfactorily during a flight is 0.985.

Let O denotes the event that original system function satisfactorily.

And let B denotes the event that backup system function satisfactorily.

Then,

  $P\left(O\right)=P\left(B\right)=0.985$

So, the probability that both systems function satisfactorily is:

  $\begin{array}{c}P\left(E\right)=P\left(O\text{ and }B\right)\\ =P\left(O\right)\times P\left(B\right)\\ =0.985\times 0.985\\ =0.9702\end{array}$

b.

To determine

To find: the probability that both systems fail.

Answer to Problem 59E

  $0.0002$

Explanation of Solution

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let $E^{\prime }$ be the event that both systems fail.

Here, backup system is independent and so the working of one system doesn’t influence that of the other system.

Also, probability that either system will function satisfactorily during a flight is 0.985.

So, the probability that either system will not work = $1-0.985=0.015$

Let $O^{\prime }$ denotes the event that original system fails.

And let $B^{\prime }$ denotes the event that backup system fails.

Then,

  $P\left(O^{\prime }\right)=P\left(B^{\prime }\right)=0.015$

So, the probability that both systems fail is:

  $\begin{array}{c}P\left(E^{\prime }\right)=P\left(O^{\prime }\text{ and }{B}^{\prime }\right)\\ =P\left(O^{\prime }\right)\times P\left(B^{\prime }\right)\\ =0.015\times 0.015\\ \approx 0.0002\end{array}$

c.

To determine

To find: the probability that both at least one system functions satisfactorily.

Answer to Problem 59E

  $\frac{841}{1600}$

Explanation of Solution

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let A be the event that at least one system functions satisfactorily.

Event A is complement of the event $E^{\prime }$ of part (b) that both systems fails.

So,

  $P\left(A\right)=1-P\left(E^{\prime }\right)$

From part (b), $P\left(E^{\prime }\right)=0.0002$

So, the probability that at least one system functions satisfactorily:

  $\begin{array}{c}P\left(A\right)=1-0.0002\\ =0.9998\end{array}$