Suppose that in a casino game the payout is a random variable X . If X is positive, you gain money, if negative, you lose.
Let p (i) = P (X = i) and suppose that:
p (0) = 1/4 ; p (1) = p (− 1) = 11/40 ; p (2) = p (− 2) = 3/32 ; and p (3) = p (− 3) = 1/160
Given that a player playing the game wins a positive amount, compute the conditional probability that..
X = 1
X = 2
X = 3

the conditional probability that X = 1 , X = 2 and X = 3 is  0.7333 (73.33%) , 0.25 (25%) and 0.0167 (1.67%) respectively

Step-by-step explanation:

a player wins money when i>0 then defining event W= gain money , then

P(W) = p(i>0) = p(1)+p(2)+p(3)

then the conditional probability can be calculated through the theorem of Bayes

P(X=1/W)= P(X=1 ∩ W)/P(W)

where

P(X=1 ∩ W)= probability that the payout is 1 and earns money

P(X=1 / W)= probability that the payout is 1 given money was earned

then

P(X=1/W)= P(X=1 ∩ W)/P(W) = P(X=1) / P(W) = p(1) /[p(1)+p(2)+p(3)] = 11/40 /(11/40+3/32+1/160 ) = 0.7333 (73.33%)

similarly

P(X=2/W)=p(2) /[p(1)+p(2)+p(3)] = 3/32 /(11/40+3/32+1/160 ) = 0.25 (25%)

P(X=3/W)=p(2) /[p(1)+p(2)+p(3)] = 1/160 /(11/40+3/32+1/160 ) = 0.0167 (1.67%)

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

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