Question

A boiler has 5 identical relief valves. The probability that any particular valve will open on demand is 0.95. An operator tries to open all five valves.

a.) Assuming independent operation of the valves, calculate the probability that at least one valve opens.

b.) Again, assuming independent operation of the valves, calculate the probability that at least one valve fails to open.

c) What is the expected number of valves that fail?

d) What is the variance of the number of values that fail?

Answer 1

Concepts and reason

Independent events: Occurrence of one event does not dependent on the occurrence of the other events.

The probability distribution of the number of successes obtained is called binomial probability distribution.

Assumptions of binomial distribution:

1)Each trail results in two exhaustive and mutually disjoint outcomes, termed as success and failure.

2)The number of trials ’n’ is finite.

3)The trials are independent of each other.

4)The probability of success ‘p’ is constant for each trial.

5)Total probability is equal to one. That is, $p + q = 1$

Fundamentals

The probability mass function of binomial distribution can be defined as,

$P\left( {X = x} \right) = \left( \begin{array}{l}\\n\\\\x\\\end{array} \right){p^x}{q^{n - x}}$

Here, $n$ be the total number of trials, $x$ be the number of trials, $p$ be the probability of success.

The mean of the binomial distribution can be defined as,

$E\left( X \right) = np$

The variance of the binomial distribution can be defined as,

${\rm{Var}}\left( X \right) = npq$

For any discrete random variable $P\left( {x \le a} \right) + P\left( {x > a} \right) = 1$

(a)

It is given that the probability of valve will open on demand is 0.95.

That is, $p = 0.95$

Valve will not open, $q = 1 - p = 1 - 0.95 = 0.05$

Number of valves, $n = 5$

Let $x$ denote that valve open.

The probability of at least one valve open is,

$\begin{array}{c}\\P\left( {x \ge 1} \right) = 1 - P\left( {x < 0} \right)\\\\ = 1 - \left\{ {\left( \begin{array}{l}\\n\\\\x\\\end{array} \right){p^x}{q^{n - x}}} \right\}\\\\ = 1 - \left\{ {\left( \begin{array}{l}\\5\\\\0\\\end{array} \right){{\left( {0.95} \right)}^0}{{\left( {0.05} \right)}^{5 - 0}}} \right\}\\\\ = 1 - {0.05^5}\\\\ = 0.9999\\\end{array}$

(b)

The probability that at least one valve fails to open is,

$\begin{array}{c}\\P\left( \begin{array}{l}\\{\rm{at least one }}\\\\{\rm{value fails to open}}\\\end{array} \right) = 1 - P\left( {{\rm{all valves open}}} \right)\\\\ = 1 - {\left( {0.95} \right)^5}\\\\ \approx 0.2262\\\end{array}$

(c)

The expected number of valves that fail is,

$\begin{array}{c}\\E\left( X \right) = nq\\\\ = 5\left( {0.05} \right)\\\\ = 0.25\\\end{array}$

(d)

The variance of valves that fail is,

$\begin{array}{c}\\{\rm{Var}}\left( X \right) = npq\\\\ = 5\left( {0.05} \right)\left( {0.95} \right)\\\\ = 0.2375\\\end{array}$

[Part d]

Part d

Ans: Part a

The probability of at least one valve open is 0.9999.

Part b

The probability that at least one valve fails to open is 0.2262.

Part c

The expected number of valves that fail is 0.25.

Part d

The variance of valves that fail is 0.2375.