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what is the binomial probability of the following: P(X < 5) P(X less than or equal to 10) P(4 < X greater than or equal to 9) n=12 p=.3 q=.7 Someone help me solve these please
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ok so the binomial probability is \[^{12}C_{r}(0.3)^r(0.7)^{12-r}\] so P(r < 5) = P(r = 0) +P(r =1) + P(r =2)+ P(r =3) + P(r = 4) Pr <= 10) = 1 - [P(r = 11) + P(r = 12)] and P(4 <= r <= 9) = P(r = 4) + P(r = 5)....+P(r = 9) hope this helps
i think it might....thank you
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