Question #332262
  1. The probability that a student must stop at any one traffic light going to NUST from Havana is 0.2. There are 15 traffic lights on the Journey.

a) what is the probability that a student will stop at exactly two of the 15 set of traffic lights?

b) what is the probability that a student will stop at 3 or more of the 15 set of traffic lights?


1
Expert's answer
2022-04-25T09:51:32-0400

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)pkqnk==(15k)0.2k0.815k==15!k!(15k)!0.2k0.815k.P(X=k)=\begin{pmatrix}n\\k\end{pmatrix}\cdot p^k\cdot q^{n-k}=\\ =\begin{pmatrix}15\\k\end{pmatrix}\cdot 0.2^k\cdot 0.8^{15-k}=\\ =\cfrac{15!}{k!\cdot(15-k)!}\cdot 0.2^k\cdot 0.8^{15-k}.


a) P(X=2)=15!2!13!0.220.813=0.2309.\text{a) } P(X=2)=\cfrac{15!}{2!\cdot13!}\cdot 0.2^{2}\cdot 0.8^{13}=0.2309.

b) P(X3)=1P(X<3)==1(P(X=0)+P(X=1)+P(X=2));P(X=0)=15!0!15!0.200.815=0.0352;P(X=1)=15!1!14!0.210.814=0.1319;P(X3)=1(0.0352+0.1319+0.2309)==0.6020.\text{b) } P(X\ge 3)=1-P(X<3)=\\ =1-(P(X=0)+P(X=1)+P(X=2));\\ P(X=0)=\cfrac{15!}{0!\cdot15!}\cdot 0.2^{0}\cdot 0.8^{15}=0.0352;\\ P(X=1)=\cfrac{15!}{1!\cdot14!}\cdot 0.2^{1}\cdot 0.8^{14}=0.1319;\\ P(X\ge 3)=1-(0.0352+0.1319+0.2309)=\\ =0.6020.



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