Question

PROBABLITY

2A. Suppose that there is a prize ceremony awarding prizes to 10 finalists. Each finalist is going to get exactly one prize, but assume that each has an equal probability of receiving any given prize. The grand prize is $5000, there are 2 prizes of $2000, 2 prizes of $1000 and 5 prizes of $200.

Let X represent the amount of money won by one prize recipient.

Find E(X), var(X) and σ for X.

2B. A pair of dice is rolled in a game.

1. Suppose a success consists of a roll with exactly one of the two dice being 6. Let X denote the number of such successes. What is the expected value of X if the dice are rolled 10 times?

2. Consider another case where a successful roll consists of either one of the dice being 6, or the total of the two being 7. Let Y denote the number of such successes. What is the least number of rolls required to achieve an expected value of at least 5 successes? (So this should be a whole number)

2C. Suppose you design a game with outcomes A, B and C for which you can estimate probabilities Pr(A) = 0.01, Pr(B) = 0.09 and Pr(C) =0.9. It costs the player $1 to play each time. If outcome C occurs, they win nothing. If outcome A occurs, they win $50. They will also win some money if B occurs. What is the largest amount paid out for B in whole dollars which will make the game break even or make a profit for you on average? Show working to justify your answer.

Answer 1

2A) The probability that a prize is won by a finalist is $\frac{1}{10}$ .

Let be thae random variable that  represents the amount of money won by one prize recipient. Then

$P\left ( X=5000 \right )=0.1\\ P\left ( X=2000 \right )=0.2\\ P\left ( X=1000 \right )=0.2\\ P\left ( X=200 \right )=0.5$

The expected value is

$E\left ( X \right )=\sum _XXP\left ( X \right )\\ E\left ( X \right )=5000\times 0.1+2000\times 0.2+1000\times 0.2+200\times 0.5\\ {\color{Blue} E\left ( X \right )=\$1200}$

Now

$E\left ( X^2 \right )=\sum _XX^2P\left ( X \right )\\ E\left ( X^2 \right )=5000^2\times 0.1+2000^2\times 0.2+1000^2\times 0.2+200^2\times 0.5\\ E\left ( X^2 \right )=3500000$

The variance is

$Var\left ( X \right )=E\left ( X^2 \right )-E\left ( X \right )^2\\ Var\left ( X \right )=3500000-1200^2\\ {\color{Blue} Var\left ( X \right )=2,080,000}$

The standard deviation is

$\sigma \left ( X \right ) =\sqrt{Var\left ( X \right )}\\ \sigma \left ( X \right ) =\sqrt{2,080,000}\\ {\color{Blue} \sigma \left ( X \right ) =\$1,442.22}$

2B)

1) The probability of success in 1 throw of 2 dice is

$\frac{10}{36}=\frac{5}{18}$

Since there are 10 outcomes that are success out of $6^2=36$ .

When the two dice are thrown 10 times the number of successes is a random variable that follws binomial distribution with $n=10,p=\frac{5}{18}$ .

The binomial probability mass function is

$P\left ( X=x \right )=\binom{n}{x}p^x\left ( 1-p \right )^{n-x},x=0,1,2,...,n$ .

The expecetd value of is $E\left (X \right )=\left ( 10 \right )\frac{5}{18}={\color{Blue} 2.7778}$ .

2) In the second case the success event is $\left \{ \left ( 1,6 \right ),\left ( 6,1 \right ),\left ( 2,5 \right ) ,\left ( 5,2 \right ),\left ( 4,3 \right ),\left ( 3,4 \right )\right \}$ .

Therefore the probability of success is $p=\frac{6}{6^2}=\frac{1}{6}$ .

As in part (1), the expected value is

$E\left (X \right )=np\geqslant 5\\ n\left ( \frac{1}{6} \right )\geqslant 5\\ n\geqslant 30$

The least number of rolls is ${\color{Blue} n= 30}$ .

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