Probability

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sathish

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Feb 27, 2012
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Hi,

Please help me with this problem.

The probability of team A winning a test match against Team B is 1/3. Both the teams plan 6 test matches.

1. What is the probability of the Team A losing all test matches
2. What is the probability of Team A winning at least one test match.

Thanks
 
sathish said:
The probability of team A winning a test match against Team B is 1/3. Both the teams plan 6 test matches.
1. What is the probability of the Team A losing all test matches
2. What is the probability of Team A winning at least one test match.
Click to expand...
We must assume that these are independent binomial trials.
Clearly the probability that B wins is \(\displaystyle 1-\tfrac{1}{3}=\tfrac{2}{3}\).

If a trial has probability \(\displaystyle p\) of success and we have \(\displaystyle N\) independent tries, then the probability of exactly \(\displaystyle K,~0\le K\le N,\) successes is \(\displaystyle \displaystyle\binom{N}{K}p^K(1-p)^{N-K}.\)

Now you tell us what is next.
 
pka said:
We must assume that these are independent binomial trials.
Clearly the probability that B wins is \(\displaystyle 1-\tfrac{1}{3}=\tfrac{2}{3}\).

If a trial has probability \(\displaystyle p\) of success and we have \(\displaystyle N\) independent tries, then the probability of exactly \(\displaystyle K,~0\le K\le N,\) successes is \(\displaystyle \displaystyle\binom{N}{K}p^K(1-p)^{N-K}.\)

Now you tell us what is next.
Click to expand...


Hi pka,

Thanks for the formula. I will try applying it and will post here.
 
Hello, sathish!

The probability of team A winning a game against Team B is 1/3.
Both the teams plan 6 games.
Click to expand...
\(\displaystyle P(A\text{ wins}) \,=\,\dfrac{1}{3}\qquad P(A\text{ loses}) \,=\,\dfrac{2}{3}\)


1. What is the probability of the Team A losing all 6 games?
Click to expand...

\(\displaystyle P(A\text{ loses 6 games}) \:=\:\left(\dfrac{2}{3}\right)^6 \:=\:\dfrac{64}{729}\)



2. What is the probability of Team A winning at least one game?
Click to expand...


\(\displaystyle P(A\text{ wins at least one game}) \:=\:1 - \dfrac{64}{729} \:=\:\dfrac{665}{729}\)
 
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