Let p be the probability of a ATM breakdown,
p=0.2
i. there are no breakdowns,
"P_1=(1-p)^5\\\\\nP_1=(0.8)^5\\\\\nP_1=0.32768"
ii. four machines break down
"P_2=(p^4 )\\times (1- p)\\\\\nP_2=(0.2)^4 \\times (0.8)\\\\\nP_2=0.00128"
iii. all machines break down
"P_3=p^5\\\\\nP_3=(0.2)^5\\\\\nP_3=0.00032"
iv. at least three machines break down
"P_4=P_2+P_3+(p^3\\times(1-p)^2)\\\\\nP_4=0.00128+0.00032+0.2^3\\times0.8^2\\\\\nP_4=0.00672"
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