There is a binomial distribution with p=0.4 and n=5.
a. P(X=0)=Cn00.40(1−0.4)5=0.0778.P(X=0)=C^0_n0.4^0(1-0.4)^5=0.0778.P(X=0)=Cn00.40(1−0.4)5=0.0778.
b. P(X≥3)=P(X=3)+P(X=4)+P(X=5)=P(X\ge3)=P(X=3)+P(X=4)+P(X=5)=P(X≥3)=P(X=3)+P(X=4)+P(X=5)=
C530.43(1−0.4)2+C540.44(1−0.4)1+C550.45(1−0.4)0=0.3174.C^3_50.4^3(1-0.4)^2+C^4_50.4^4(1-0.4)^1+C^5_50.4^5(1-0.4)^0=0.3174.C530.43(1−0.4)2+C540.44(1−0.4)1+C550.45(1−0.4)0=0.3174.
c. E(X)=pn=0.4∗5=2.E(X)=pn=0.4*5=2.E(X)=pn=0.4∗5=2.
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