Consider an urn initially containing 6 amber balls and 2 blue balls. Player A wi
ID: 3047245 • Letter: C
Question
Consider an urn initially containing 6 amber balls and 2 blue balls. Player A will draw balls from this urn, one after another, until he obtains a blue ball, and his score will be the number of balls he has to draw in order to obtain a blue ball. Next, without putting any balls back in the urn, Player B will start to drawn balls from this urn, one after another, until he obtains the second blue ball, and his score will be the number of balls he has to draw in order to obtain the second blue ball. What is the probability that Player B’s score will be greater than Player A’s score?
Explanation / Answer
Lets say X is the number of draws required for A and Y is number of draws required by B.
If X = 1, Y can have values from 1 to 7
if X = 2, Y can have values from 1 to 6
if X =3 , Y can have values form 1 to 5
if X = 4, Y can have values from 1 to 4
if X = 5, Y can have values form 1 to 3
if X = 6, Y can values from 1 to 2
if X = 7, Y can have value 1 only.
Now only X = 1, 2,3 are good here for which player B's score will be greater than player A's score.
so if X = 1, means first ball drawn is blue one. so to be Y greater than 1 , in the remaining seven balls, except first chance if we get blue ball in any other chance we will get Y greater than X.
so Pr(Y > X l X = 1) = Pr(Y = (6/7) * (2/8) = 6/7 * 1/4 = 6/28 = 3/14
Pr(Y > X l X = 2) = Pr(Y = 3,4,5,6) = Pr(Amber ball in first chance) * Pr(BLue ball in second draw by player A) * Pr(Not blue ball in first two draw) = (6/8) * (2/7) * (5/6) * (4/5) = 1/7
Pr(Y >X l X= 3) = Pr(Y = 4,5) = Pr(AMber in first chance) * Pr(Amber in second chance) * Pr(Blue in third draw by player A) * Pr(not blue ball in initial three draw)
= (6/8) * (5/7) * (2/6) * (4/5) * (3/4) * (2/3) = 1/14
Pr(B score is greater than A score) = 3/14 + 1/7 + 1/14 = 2/7 + 1/7 = 3/7
so the probability is 3/7