Question #156625

Select three fans randomly at a football game in which Penn State is playing over Notre Dame. Identify whether the fan is Penn State fan (P) or a Notre Dame fan (D). Let X be the number of Penn State fans.


1
Expert's answer
2021-01-20T13:10:21-0500

The fan is a Penn State fan (P)(P) or a Notre Dame fan (N).(N). This experiment yields the following sample space:

S={PPP,PPN,PNP,PNN,NPP,NPN,NNP,NNN}S=\{PPP, PPN,PNP, PNN, NPP, NPN, NNP, NNN\}

Let X=X= the number of Penn State fans selected. The possible values of XX are, therefore, either 0,1,2,0,1,2, or 3.3.

Since the game is a home game, let's suppose that 80% of the fans attending the game are Penn State fans, while 20% are Notre Dame fans. That is, P(P)=0.8P(P)=0.8 and P(N)=0.2P(N)=0.2

Then, by independence:


P(X=0)=P(NNN)=(0.2)3=0.008P(X=0)=P(NNN)=(0.2)^3=0.008

By independence and mutual exclusivity of NNP,NPN,NNP, NPN, and PNNPNN


P(X=1)=P(NNP)+P(NPN)+P(PNN)P(X=1)=P(NNP)+P(NPN)+P(PNN)

=3(0.8)(0.2)2=0.096=3(0.8)(0.2)^2=0.096

By independence and mutual exclusivity of PPN,PNP,PPN, PNP, and NPPNPP


P(X=2)=P(PPN)+P(PNP)+P(NPP)P(X=2)=P(PPN)+P(PNP)+P(NPP)

=3(0.8)2(0.2)=0.384=3(0.8)^2(0.2)=0.384

By independence:


P(X=3)=P(PPP)=(0.8)3=0.512P(X=3)=P(PPP)=(0.8)^3=0.512


Check


0.008+0.096+0.384+0.512=10.008+0.096+0.384+0.512=1

The probability of the sample space is 1.1.

Because the values that it takes on are random, the variable XX is a random variable (rv).




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