Question

Niyat is a gambler and regularly plays several rounds of a gamble in which he wins $3,000 if even number of dots face up when a fair die is rolled twice and loses $1,000 for any other outcome from the two rolls of the die. In other words, the outcome of the gamble is determined by rolling a die twice on each round of the gamble and Niyat wins only if even number of dots show up on top in both rolls of the die and loses if any other outcome occurs. Niyat is considering playing 25 rounds of such a gamble next week hoping that she will win back the $1,500 she lost in a similar gamble last week. If we let X represent the number of wins for Niyat out of the next 25 rounds of the gamble, X will have the binomial probability distribution. a. Please calculate the probability of success for Niyat on each round of the gamble. Show how you arrived at your answer. b. What is the probability that Niyat will lose none of the 25 rounds of the gamble? c. What is the probability that Niyat will win more than half of the 25 rounds of the gamble? Show your work d. What is the probability that Niyat will win at least 8 out of the 25 rounds of the gamble? Show your work. e. What is the probability that Niyat will lose less than 20 of the 25 rounds of the gamble? Show your work. f. What is the probability that Niyat will lose at most 15 out of the 25 rounds of the gamble? Show your work g. How much money is Niyat expected to win in the 25 rounds of the gamble? How much money is she expected to lose? Given your results, do you think Niyat is playing a smart gamble? Please show how you arrived at your results and explain your final answer. h. Calculate and interpret the standard deviation for the number of wins for Niyat in the next 25 rounds of the gamble. Show your work.

Answer 1

When two dice are rolled the possible sample space showing the total number of total dots on two faces of dice:

Out of 36 outcomes, 9 shows odd number of dots on both dice. So probability of winning is single trial is

P(win) = 9/36 = 0.25

Here X has binomial distribtuion with parameters n=11 and p = 0.25. The pdf of X is

Following table shows the required probability :

X	P(X=x)
0	0.04224
1	0.15486
2	0.2581
3	0.2581
4	0.17207
5	0.0803
6	0.02677
7	0.00637
8	0.00106
9	0.00012
10	0.00001
11	0.00000

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The probability that Joe will win none of the 11 rounds of the gamble is

P(X= 0 ) = 0.04224

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the probability that Joe will win more than 6 of the 11 rounds of the gamble is

P(X > 6) = 0.00756

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The probability that Joe will lose not more than 3 out of the 11 rounds of the gamble is

P(X >= 8) = 0.00119

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The probability that Joe will lose at least 5 the 11 rounds of the gamble is

P(X <= 6) = 0.99244

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The probability that Joe will lose fewer than 9 out of the 11 rounds of the gamble is

P(X >= 3) = 0.5448