Question 1021899
So find the probability that 2 bombs hit the target.
{{{P(2)=C(6,2)(0.2)^2(1-0.2)^(6-2)}}}
{{{P(2)=15(4/100)(4096/10000)}}}
{{{P(2)=24576/100000}}}
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So find the probability that 4, 5, and 6 candidates passed.
{{{P(4)=C(6,4)(0.6)^4(0.4)^(6-4)=15(1296/10000)(16/10000)=31104/100000}}}
{{{P(5)=C(6,5)(0.6)^5(0.4)^(6-5)=6(7776/100000)(4/10)=186624/1000000}}}
{{{P(6)=C(6,6)(0.6)^6(0.4)^(6-6)=1(46656/1000000)(1)=46656/1000000}}}
So then,
{{{P=P(4)+P(5)+P(6)}}}
{{{P=(311040+186624+46656)/1000000}}}
{{{P=54432/100000}}}