We have a Bernoulli trial - exactly two possible outcomes, "success" (the student stops at a traffic light) and "failure" (he doesn't stop) and the probability of success is the same every time the experiment is conducted (the student mets a traffic light).
The probability of each result
P(X=k)=(nk)⋅pk⋅qn−k==(15k)⋅0.2k⋅0.815−k==k!⋅(15−k)!15!⋅0.2k⋅0.815−k.
a) P(X=2)=2!⋅13!15!⋅0.22⋅0.813=0.2309.
b) P(X≥3)=1−P(X<3)==1−(P(X=0)+P(X=1)+P(X=2));P(X=0)=0!⋅15!15!⋅0.20⋅0.815=0.0352;P(X=1)=1!⋅14!15!⋅0.21⋅0.814=0.1319;P(X≥3)=1−(0.0352+0.1319+0.2309)==0.6020.
Comments