Let "X=" the number of machines which are break down: "X\\sim Bin (n;p)"
"P(X=x)=\\binom{n}{x}p^x(1-p)^{n-x}" Given that "n=5, p=0.2"
i. there are no breakdowns
"P(X=0)=\\binom{5}{0}(0.2)^0(1-0.2)^{5-0}=""=(0.8)^5=0.32768"
ii. four machines break down
"P(X=4)=\\binom{5}{4}(0.2)^4(1-0.2)^{5-4}=""=5(0.2)^4 (0.8)=0.0064" iii. all machines break down
"P(X=5)=\\binom{5}{5}(0.2)^5(1-0.2)^{5-5}=""=(0.2)^5=0.00032" iv. at least three machines break down
"P(X\\geq 3)=P(X=3)+P(X=4)+P(X=5)=""=\\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"
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