Answer:
a) P(4 successes) = 0.127
b) P(2 successes) = 0.283
c) P(at most 3 successes) = 0.984
Explanation:
The binomial distribution formula is given as:
[tex]P_x=nCx(p^x)(q^{n-x})[/tex]
Note that:
[tex]nCx=\frac{n!}{(n-x)!x!}[/tex]
This combination formula will be used to calculate all combination expressions in this exercise.
a) For n = 15, p = 0.4
q = 1 - p = 1 - 0.4 = 0.6
[tex]\begin{gathered} P(x=4)=(15C4)(0.4^4)(0.6^{15-4}) \\ \\ P(x=4)=1365(0.0256)(0.00362797056) \\ \\ P(x=4)=0.127 \end{gathered}[/tex]
b) For 12, p = 0.2
q = 1 - 0.2 = 0.8
[tex]\begin{gathered} P(x=2)=(12C2)(0.2^2)(0.8^{12-2}) \\ \\ P(x=2)=66(0.04)(0.1073741824) \\ \\ P(x=2)=0.283 \end{gathered}[/tex]
c) n = 20, p = 0.05
q = 1 - 0.05 = 0.95
P(at most 3 successes) = P(x=0) + P(x=1) + P(x = 2) + P(x=3)
[tex]\begin{gathered} P(x=0)=(20C0)(0.05)^0(0.95)^{20}=0.358 \\ P(x=1)=(20C1)(0.05)^1(0.95)^{19}=0.377 \\ P(x=2)=(20C2)(0.05)^2(0.95)^{18}=0.189 \\ P(x=3)=(20C3)(0.05)^3(0.95)^{17}=0.060 \\ \end{gathered}[/tex]
P(at most 3 successes) = 0.358 + 0.377 + 0.189 + 0.060
P(at most 3 successes) = 0.984
