# Probability

• Oct 4th 2008, 05:44 AM
santatrue
Probability
There are two players playing named John and Peter. The one who wins three games first is the winner of the match. So the max no. of games needed is 5. And John has probability of winning is 0.6 and assume that all games are independent.

(a)Given that John has lost the first two games, Find the probability so that he can still be the winner of the match?
(b) Given that John lost the first game, Find the probability that John will still be the winner?
(c) Find the probability that the number of games needed is exactly 5?

Sorry for so many questions but I have tried a lot and am unable to answer!
• Oct 4th 2008, 07:30 AM
Plato
Quote:

Originally Posted by santatrue
There are two players playing named John and Peter. The one who wins three games first is the winner of the match. So the max no. of games needed is 5. And John has probability of winning is 0.6 and assume that all games are independent.
(a)Given that John has lost the first two games, Find the probability so that he can still be the winner of the match?
(b) Given that John lost the first game, Find the probability that John will still be the winner?
(c) Find the probability that the number of games needed is exactly 5?

For part (a), what is the probability of this string, PPJJJ?

For part (b), what is the probability of each of these strings, PJJJ, PPJJJ, PJPJJ, PJJPJ? [Add these four.]

For part (c), what is the probability of each of these strings, PPJJJ, PJPJJ, PJJPJ, JPPJJ, JPJPJ, JJPPJ, JJPPP, JPJPP, JPPJP, PJJPP, PPJJP, PJPJP? [Add these 12.]
• Oct 4th 2008, 07:44 AM
santatrue
Quote:

Originally Posted by Plato
For part (a), what is the probability of this string, PPJJJ?

For part (b), what is the probability of each of these strings, PJJJ, PPJJJ, PJPJJ, PJJPJ? [Add these four.]

For part (c), what is the probability of each of these strings, PPJJJ, PJPJJ, PJJPJ, JPPJJ, JPJPJ, JJPPJ, JJPPP, JPJPP, JPPJP, PJJPP, PPJJP, PJPJP? [Add these 12.]

isnt is suppose to be Conditional Probability or you can also do this way
• Oct 4th 2008, 07:52 AM
Plato
Quote:

Originally Posted by santatrue
isnt is suppose to be Conditional Probability

I used Conditional Probability!
Given John looses the first two means that Peter wins the first two.
Look at it this way, Conditional Probability just restricts our attention to a subset of the outcome space.
• Oct 4th 2008, 07:56 AM
kbartlett
Quote:

Originally Posted by santatrue
There are two players playing named John and Peter. The one who wins three games first is the winner of the match. So the max no. of games needed is 5. And John has probability of winning is 0.6 and assume that all games are independent.

(a)Given that John has lost the first two games, Find the probability so that he can still be the winner of the match?
(b) Given that John lost the first game, Find the probability that John will still be the winner?
(c) Find the probability that the number of games needed is exactly 5?

Sorry for so many questions but I have tried a lot and am unable to answer!

(a) The Propability is $\displaystyle \frac{PPJJJ}{PP}$ = $\displaystyle JJJ$

(b) $\displaystyle JJJPx4$ = $\displaystyle 4(0.6)^3(0.4)$

(c) $\displaystyle 6(0.6)^3(0.4)^2+6(0.6)^2(0.4)^3$
• Oct 4th 2008, 07:56 AM
santatrue
Quote:

Originally Posted by Plato
I used Conditional Probability!
Given John looses the first two means that Peter wins the first two.
Look at it this way, Conditional Probability just restricts our attention to a subset of the outcome space.

So shouldnt it be for the first question to find for JJJ only , why PP as he has already won
• Oct 4th 2008, 08:11 AM
kbartlett
Quote:

Originally Posted by santatrue
So shouldnt it be for the first question to find for JJJ only , why PP as he has already won

He was Wrong
If
A = The Probability that Paul wins the first 2 games.
B = The probaility that John wins overall.

Then the probability of B given A is:
P(B|A) = $\displaystyle \frac{P(A and B)}{P(A)}$= $\displaystyle \frac{PPJJJ}{PP}$ = $\displaystyle \frac{(0.4)^2(0.6)^3}{(0.4)^2}$ = $\displaystyle 0.6^3$
• Oct 4th 2008, 08:33 AM
Plato
Quote:

Originally Posted by santatrue
So shouldnt it be for the first question to find for JJJ only , why PP as he has already won

Sorry that I did not make it clear that you do indeed divide by the given probability. I see how I misslead you.