Question

Players A and B each roll a fair 6-sided die. The player with the higher score wins ¤1 from the other player. If both players have equal scores, the game is a draw and no one wins anything.

i. Let X denote the winnings of player A from one round of this game. State the probability mass function of X. Calculate the expectation E(X) and variance Var(X).

ii. What is the conditional probability that player A rolls , given that player B wins. iii. Suppose the game is repeated 10 times. Compute the probability that at least two of the rounds result in draws. [20 marks]

(b) i. The number of times that a person contracts a cold in a given year is a Poisson random variable with parameter λ = 5. Calculate the probability that a person contracts 2 colds in a year.

ii. Suppose that a new wonder drug (based on large quantities of vitamin C) has just been marketed that reduces the Poisson parameter to λ = 3 for 75% of the population. For the other 25% of the population, the drug has no effect on colds. If an individual tries the drug for a year and has 2 colds in that time, what is the probability that the drug is beneficial for him or her?

Answer 1

$i.~Let~(a,b)~denotes~the~outcome~where~"a"~is~obtained~by~A~and~"b"~is~obtained~by~B\\ where,~a,b=1(1)6\\ P(X=1)=P(\{(2,1),(3,1),(3,2),(4,3),(4,2),(4,1),(5,4),(5,3),(5,2),(5,1),\\ (6,5),(6,4),(6,3),(6,2),(6,1)\})=\frac{15}{36}\\ P(X=0)=1-\frac{15}{36}=\frac{21}{36}\\ E(X)=\frac{15}{36}=\frac{3}{12},~E(X^2)=\frac{15}{36},~Var(X)=\frac{15\times 21}{36^2}=\frac{35}{144}\\ ii.~Let~U~be~a~random~variable~denoting~the~no.~is~obtained~by~A\\ B~wins\Rightarrow \{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\}\\ then~P(U=1|B~wins)=\frac{5}{15},~P(U=2|B~wins)=\frac{4}{15},\\ ~P(U=3|B~wins)=\frac{3}{15},~P(U=4|B~wins)=\frac{2}{15},\\ ~P(U=5|B~wins)=\frac{1}{15}\\ iii.~Let~X~be~random~variable~denote~no.~of~draws~occurred.\\ Then~X\sim Binomial(10,6/36=1/6)~since~draws~occurred~when\\ any~one~of~the~following~outcomes~occurs:\\ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6).\\ Required~probability=P(X\geq 2)=1-P(X=0)-P(X=1)\\ =1-(5/6)^{10}-10*(1/6)*(5/6)^9=0.5155\\$

$(b)~i.~P(X=2)=e^{-5}5^2/2!=0.0842\\ (c)~Let~E~be~an~event~that~the~drug~is~effective~and~X~be~an~random~variable~denoting~\\ the~number~of~times~that~a~person~contracts~a~cold~and~X\sim Poi(3).\\ The~required~probability=P(E|X=2)=\frac{P(X=2|E)P(E)}{P(X=2|E)P(E)+P(X=2|E^c)P(E^c)}\\ =\frac{(e^{-3}3^2/2!)*0.75}{(e^{-3}3^2/2!)*0.75+(e^{-3}3^2/2!)*0.25}=0.75$