Question

Problem 3. During breaks in a basketball game, fans can enter a free throw contest against the team mascot, who is a former player with an 85% career free-throw percentage. The fans' basketball skills are more modest: an average fan has a 45% chance of making a free throw. The fan and the mascot each take one shot (ties are possible). a) What is the probability that the fan wins? b) What is the probability that the mascot wins? c) In case of a tie, what is the probability both the fan and the mascot make their shots? d) The prize for winning a contest is $400, the prize for tying is $25, and there is no prize for losing. What is the mean prize and the standard deviation of the prize? e) During one season, 286 fans were selected from the audience at random to participate in the contest. What is the distribution of the average prize for this sample? What is the probability the average prize exceeded $50?

Answer 1

3.

Suppose,

M represents the event of successful throw by the mascot.

F represents the event of successful throw by a fan.

So, P(M) = 0.85 and P(F) = 0.45

Clearly, M and F are two mutually independent events.

(a)

Probability of a fan to win = P(Fan wins)

P(F) P(M)

= 0.45*(1-0.85) = 0.0675

(b)

Probability of a mascot to win = P(Mascot wins)

P(M) P(F)

= 0.85*(1-0.45) = 0.4675

(c)

This is a problem of conditional probability.

Probability of tie = P(Tie)

P(M) P(F) P(M) P(F)

= 0.85*0.45 + (1-0.85)*(1-0.45) = 0.465

In case of tie, probability of both make their shots = P(Both make shots | Tie occurred)

[Since, both F and M can occure in case of tie only]

P(FnM)nTie) P((FnM) P(Tie) P(Tie)

[Since, F and M are independent]

P(F) P(M) Р(Tie)

0.45 0.85 0.465

= 0.8226

(d)

Suppose,

W denotes the event of winning of a fan.

X denotes amount won by a fan.

So, P(W) = 0.0675

E(X) 400 P(W)25 P(W)

= 400*0.0675+25*(1-0.0675) = 50.3125

So, mean prize = $50.31

Е(X) — * Р(W) 4002 P(W) +252

= 11382.8125

- E(X) Var (X) E(Xx2)

   = 8851.4648

So, standard deviation of prize = $94.08