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?


Expert's answer

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==15!k!⋅(15−k)!⋅0.2k⋅0.815−k.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!⋅0.22⋅0.813=0.2309.\text{a) } P(X=2)=\cfrac{15!}{2!\cdot13!}\cdot 0.2^{2}\cdot 0.8^{13}=0.2309.

b) P(X≥3)=1−P(X<3)==1−(P(X=0)+P(X=1)+P(X=2));P(X=0)=15!0!⋅15!⋅0.20⋅0.815=0.0352;P(X=1)=15!1!⋅14!⋅0.21⋅0.814=0.1319;P(X≥3)=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.



LATEST TUTORIALS
APPROVED BY CLIENTS