Suppose that in a casino game the payout is a random...

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) = 1/3; p(1) = p(-1) = 49/195; p(2) = p(-2) = 1/13; and p(3) = p(-3) = 1/195. Given that a player playing the game wins a positive amount, compute the conditional probability that... ... X = 1: ... X = 2: ... X = 3:

Transcript

00:01 In this question, we're given that x is the payout of a casino game.

00:06 And if x is positive, you gain money, x less than zero, you lose money.

00:12 We're given that the small p -i is actually the probability of x equals to i.

00:18 And we're given that p -0 is wanted.

00:21 P -1 is 49 -5.

00:24 P -2 is 1 over -13, and p -3 is 1 over -195.

00:29 We want to find the probability x equals to 1 given the player playing the game wins a positive amount so what we are looking at is we want to find the probability of x equals to 1 given x is greater and 0 because when x squared is 0 he wins with a positive amount so by bale's theorem this is probability x equals to 1 intersect with x greater than 0 over probability x greater than 0.

01:03 Now let's look at probability x squared and zero.

01:06 Over here, probability x greater than 0 will be probability of 1 plus probability of 2 plus probability of 3.

01:20 So this will be adding up all this probability.

01:23 Now why do i choose plus and not times? because these are mutually exclusive events.

01:31 So it's either p1 or now all is plus for probability all is plus n is times so it's either this case or so plus plus this case or this case so all are plus so adding up all this you will get one third all right so over back to here the numerator probability x equals 1 intersect x squared 0 and so that's just x equals to 1.

02:05 So probability of s equals to 1 is this...

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