Answer to Question #155582 in Statistics and Probability for abd el rahmm

Question

A machine operates without a breakdown in 90% of the times on a given day. What is the probability that in one week of 6 working days there are at most 3 breakdowns that occur?

Select one:

0.9841

0.0159

0.9987

0.0013

Clear my choice

Answer 1

Here we have that

P(machine operates without a breakdown) = 0.9

Hence P(breakdown) = 1 – 0.9 = 0.1

n = 6

x = 3

The number of breakdowns that occur among a fixed number of trials follows the binomial distribution.

The binomial distribution formula:

$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{(n-x)}$

The probability that in one week of 6 working days there are at most 3 breakdowns that occur:

$P(X\le3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=0)=C(6,0)\cdot0.1^0\cdot0.9^{6-0}=0.5314$

$P(X=1)=C(6,1)\cdot0.1^1\cdot0.9^{6-1}=0.3543$

$P(X=2)=C(6,2)\cdot0.1^2\cdot0.9^{6-2}=0.0984$

$P(X=3)=C(6,3)\cdot0.1^3\cdot0.9^{6-3}=0.0146$

$P(X\le3)=0.5314+0.3543+0.0984+0.0146=0.9987$

Answer:

The probability that in one week of 6 working days there are at most 3 breakdowns that occur is 0.9987