132/3*2/3+1/3*2/3*2*2/3*2/3+1/3*2/3*2*1/3*2/3*2*2/3*2/3+..........=4/5.....got it....thanx nishant bhiya
i missied the case:Deuce -> B wins -> A wins (Back to deuce) P = 2/3 x 1/3
A and B plays a game of tennis. The situation of the game is as follows: if one scores two consecutive points after a deuce he wins; if loss of a point is followed by win of a point, it is deuce. The chance of a server to win is 2/3. The game is at deuce and A is serving. Find the probability that A wins.
edit: serve changes after each turn
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11 Answers
33No....
:( i got 2/5 but answer is neither of the two
and i am not finding any mistake in either my method or that of the package..
33No...
:(
Ok options are (for tukka :P)
a) 3/5
b)2/5
c)1/2
d)4/5
Both mine and Nishant bhaiya's answers were there in option so definitely both of us will have done it wrong in xam
:P
13
62I will tell u the method..
I am right :P
:D
seee the cases
Deuce - > A wins -> A wins P=2/2 x 2/3
Deuce -> A wins -> B wins (Back to deuce) P=1/3 x 2/3
Deuce -> B wins -> A wins (Back to deuce) P = 2/3 x 1/3
Deuce -> B wins -> B wins
Total probabiltiy of A winning from a deuce =P(A) let
hence from the above
P(A) = 2/3x2/3 + 1/3x2/3xP(A) + 2/3x1/3xP(A) + 0
Get P(A) = 4/5
33Opsie a part of question i missed
(edited in the question)
Serve changes after each point..
:P
62for the new edited question
seee the cases
Deuce - > A wins -> A wins P=2/3 x 1/3
Deuce -> A wins -> B wins (Back to deuce) P=2/3 x 2/3
Deuce -> B wins -> A wins (Back to deuce) P = 1/3 x 2/3
Deuce -> B wins -> B wins
Total probabiltiy of A winning from a deuce =P(A) let
hence from the above
P(A) = 2/3x1/3 + 2/3x2/3xP(A) + 1/3x1/3xP(A) + 0
Get P(A) = 1/2
132/3*1/3+[(2/3*2/3+1/3*1/3)2/3*1/3]+......=2/9[1/(1-5/9)]=1/2
33I did the same but missed the term
P(A) = 2/3x1/3 + 2/3x2/3xP(A) + 1/3x1/3xP(A) + 0
:(
Let me see which term is it?
33Got it
:)
In b/w The soln given (for others)
Just try to convince yourself what is being done :)
Req prob=
Σ(5/9)r(2/3)(1/3) .... summation from 0 to ∞
:)