0.2158Chris tries to throw a ball of paper in the wastebasket behind his back (w

Question

0.2158Chris tries to throw a ball of paper in the wastebasket behind his back (without looking). He estimates that his chance of success each time, regardle of the outcome of the other attempts, is 0.32. Let X be the number of attemp required to get the ball of paper into the wastebasket. If he is not successful within the first 5 attempts, then he quits and he lets X = 6 in such a case. Y indicate whher he nakosth baskoi succsully within the firsihr attempts. Thus Y = 1 if his first, second, or third attempt is successful, and Y. 0 otherwise. Find the conditional mass of X given Y, that is, find P(X = x | Y = y). Give all probabilities to four decimal places (0.0001) yl= 0.685568

Explanation / Answer

WHen Y = 1, then X = 1,2 and 3

when Y = 0 then X = 4, 5 and 6

Pr( Success in any given attempt) = 0.32

for X = 1, Pr(X =1) = 0.32

for X = 2, Pr(X =2) = 0.32 * 0.68 = 0.2176

for X = 3, Pr(X = 3) = 0.32 * 0.682 = 0.1480

for X = 4 , Pr(X = 4) = 0.32 * 0.683 = 0.1006

for X = 5, Pr( X = 5) = 0.32 * 0.684 = 0.0684

for X = 6, Pr (X = 6) = 1 - [Pr(0) + Pr(1) + Pr(2) + Pr(3) + Pr(4) + Pr(5)] = 0.1454

so P(X =x l Y= y)

P(X =1 l Y =1) = 0.32 / ( 0.32 + 0.2176 + 0.1480) = 0.4667

P(X =2 l Y = 1) = 0.2176/ (0.32 + 0.2176 + 0.1480) = 0.3174

P(X = 3 l Y =1) = 0.1480 / (0.32 + 0.2176 + 0.1480) = 0.2159

P( X = 4,5,6 l Y =1) = 0

P(X =4 l Y = 0) = 0.1006 / ( 0.1006 + 0.0684 + 0.1454) = 0.32

P(X =5 l Y = 0) = 0.0684/ ( 0.1006 + 0.0684 + 0.1454) = 0.2176

P(X =6 l Y =0) = 0.1454 / ( 0.1006 + 0.0684 + 0.1454) = 0.4624

P ( X = 1,2,3 l Y = 0) = 0

Y/X X = 1 X = 2 X = 3 X = 4 X = 5 X = 6 Y= 0 0 0 0 0.32 0.2176 0.4624 Y= 1 0.4667 0.3174 0.2159 0 0 0

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