Question

g The probability that a machine breaks down in any day is 0.20 and is independent of the breakdown on any other day. The machine can break down only once per day. Calculate the probability that the machine breaks down two or more times in ten days.

Answer 1

Answer:

62.42% probability that the machine breaks down two or more times in ten days.

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the machine breaks, or it does not break. The probability of the machine breaking in a day is independent from other days. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that a machine breaks down in any day is 0.20

This means that $p = 0.2$

Calculate the probability that the machine breaks down two or more times in ten days.

This is $P(X \geq 2)$ when n = 10.

We know that either the machine breaks in less than two days, or it breaks in at least two days. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

$P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.1074 + 0.2684 = 0.3758$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.3758 = 0.6242$

62.42% probability that the machine breaks down two or more times in ten days.

Answer 2

Answer:

62.42%

Step-by-step explanation: