Question #107931
. There are five old ATM machines at Smart Bank. All probability that an ATM breaks
down in the course of a day is 0.2. If they are all in working order at the beginning a
day, and breakdowns are independent of each other, find the probability that during
the day;
i. there are no breakdowns
ii. four machines break down
iii. all machines break down
iv. at least three machines break down
1
Expert's answer
2020-04-06T15:53:48-0400

Let X=X= the number of machines which are break down: XBin(n;p)X\sim Bin (n;p)


P(X=x)=(nx)px(1p)nxP(X=x)=\binom{n}{x}p^x(1-p)^{n-x}

Given that n=5,p=0.2n=5, p=0.2

i. there are no breakdowns 


P(X=0)=(50)(0.2)0(10.2)50=P(X=0)=\binom{5}{0}(0.2)^0(1-0.2)^{5-0}==(0.8)5=0.32768=(0.8)^5=0.32768

ii. four machines break down 


P(X=4)=(54)(0.2)4(10.2)54=P(X=4)=\binom{5}{4}(0.2)^4(1-0.2)^{5-4}==5(0.2)4(0.8)=0.0064=5(0.2)^4 (0.8)=0.0064

iii. all machines break down 


P(X=5)=(55)(0.2)5(10.2)55=P(X=5)=\binom{5}{5}(0.2)^5(1-0.2)^{5-5}==(0.2)5=0.00032=(0.2)^5=0.00032

iv. at least three machines break down


P(X3)=P(X=3)+P(X=4)+P(X=5)=P(X\geq 3)=P(X=3)+P(X=4)+P(X=5)==(53)(0.2)3(10.2)53+0.0064+0.00032==\binom{5}{3}(0.2)^3(1-0.2)^{5-3}+0.0064+0.00032==10(0.2)3(0.8)2+0.00672=0.05792=10(0.2)^3(0.8)^2 +0.00672=0.05792


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