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

Question #155582

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




1
Expert's answer
2021-01-14T20:03:02-0500

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)px(1p)(nx)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(X3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)P(X\le3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)

P(X=0)=C(6,0)0.100.960=0.5314P(X=0)=C(6,0)\cdot0.1^0\cdot0.9^{6-0}=0.5314

P(X=1)=C(6,1)0.110.961=0.3543P(X=1)=C(6,1)\cdot0.1^1\cdot0.9^{6-1}=0.3543

P(X=2)=C(6,2)0.120.962=0.0984P(X=2)=C(6,2)\cdot0.1^2\cdot0.9^{6-2}=0.0984

P(X=3)=C(6,3)0.130.963=0.0146P(X=3)=C(6,3)\cdot0.1^3\cdot0.9^{6-3}=0.0146

P(X3)=0.5314+0.3543+0.0984+0.0146=0.9987P(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


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