Question

Will rate!!

Answer 1

a) The highest value of P(x) is 0.208 for when X=$300. Hence that is the most likely dollar amount that can be won.

ans: The most likely dollar amount is $300

b) When you spin Lose a turn or Bankrupt, the amount won is x=$0. The probability value P(x) for x=0 is 0.083.

ans: The probability to spin Lose a turn or Bankrupt is 0.083

c) the average dollar amount is the expected value of X. It is calculated as below

$\begin{align*} E(X)&=\sum_xxP(x)\\ &=0\times P(0)+ 300\times P(300)+ 350\times P(350)+ 400\times P(400)+ 450\times P(450)+ 500\times P(500)+ 550\times P(550)+ 600\times P(600)+ 700\times P(700)+ 800\times P(800)+ 900\times P(900)+ 5000\times P(5000)\\ &=0\times 0.083+ 300\times 0.208+ 350\times 0.042+ 400\times 0.083+ 450\times 0.042+ 500\times 0.125+ 550\times 0.042+ 600\times 0.125+ 700\times 0.042+ 800\times 0.083+ 900\times 0.083+ 5000\times 0.042\\ &=668.75 \end{align*}$

The average dollar amount is $668.75

d) The probability that the contestant spins $0 in any given turn is P(x=0)=0.083

the probability that the contestant spin $0 each time is

(Probability that spin $0 first time)x(Probability that spin $0 second time)x(Probability that spin $0 third time) =

P(X=0)xP(X=0)xP(X=0) = 0.083*0.083*0.083 = 0.0006.

We can also solve this using a Binomial distribution as given below.

Let Y be the number of times the contestant spins $0 out of 3 spins. We can say that y has a Binomial distribution with parameters, number of trials n=3 and the success probability p=0.083.

The probability of spinning $0, Y=y number of times is given by (the Binomial probability is)

$\begin{align*} P(Y=y)&=\binom{n}{y}p^y(1-p)^{n-y}\\ &=\binom{3}{y}0.083^y(1-0.083)^{3-y}\\ &=\frac{3!}{y!(3-y)!}0.083^y(1-0.083)^{3-y}\\ \end{align*}$

The probability that the contestant spin $0, 3 out of 3 times is given by the probability of Y=3

$\begin{align*} P(Y=3)&=\binom{3}{3}0.083^3(1-0.083)^{3-3}\\ &=\frac{3!}{3!(3-3)!}0.083^3(1-0.083)^{0}\\ &=0.083^3\\ &=0.0006 \end{align*}$

ans: The probability that the contestant spin $0, 3 out of 3 times is 0.0006

e) The probability that the contestant spins $0 at least one time out of 3 is same as 1- (probability that the contestant does not spin $0 each of 3 times)

The probability that the contestant does not spin $0 in any given turn is 1-P(x=0)=(1-0.083) = 0.917

the probability that the contestant spin does not $0 each of 3 times is

(Probability that does not spin $0 first time)x(Probability that does not spin $0 second time)x(Probability that does not spin $0 third time) =(1-P(X=0))x(1-P(X=0))x(1-P(X=0)) = 0.917*0.917*0.917 = 0.7703

The probability that the contestant spins $0 at least one time out of 3 is same as 1- (probability that the contestant does not spin $0 all 3 times) = 1-0.7703=0.2297

Alternatively, we can use Binomial distribution.

The probability that the contestant spin $0 at least one time out of 3 is same as the probability that Y is 1 or more

$\begin{align*} P(Y\ge 1)&=1-P(Y<1)\\ &=1-P(Y=0)\\ &=1-\binom{3}{0}0.083^0(1-0.083)^{3-0}\\ &=1- \frac{3!}{0!(3-0)!}0.083^0(1-0.083)^{3}\\ &=1-(1-0.083)^{3}\\ &=1-0.7703\\ &=0.2297 \end{align*}$