SOLUTION: Was wondering if someone could give me the probability and the equation to solve the following. Two teams are to play four games against each other. In each game Team A is 45% t

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Question 850137: Was wondering if someone could give me the probability and the equation to solve the following.
Two teams are to play four games against each other. In each game Team A is 45% to win and team B is 55% to win. What it the probability of Team A winning 2 or more of the games (i.e. 2,3 or 4)? Also if you can show me how to figure it out that would be great.
Thanks

Found 2 solutions by ewatrrr, rothauserc:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
Note: The probability of x successes in n trials is:
P (x)= (p)^x(q)^(n-x) where p and q are the probabilities of success and failure respectively.
In this case p(A wins) = .45 & q(A loses) = .55 n = 4
P(x= 2) = (.45)^2(.55)^2
P(x= 3) = (.45)^3(.55)^1
P(x= 4) = (.45)^4(.55)^0
P(x ≥ 2)= P(x= 2) + P(x= 3) + P(x= 4)
0r using TI Calculator
p = .05, n = 100
P(x ≥ 2) = 1 – P(x ≤ 1) = 1 - binomcdf(4, .45, 1)



Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
Determine the probability of Team A winning 0 or 1 game, then we can subtract from 1.
p = 0.45 (Team A wins, B loses)
q = 0.55 (Team B wins, A loses)
For 0 wins, that would be the probability of 4 wins by team B.
P(x=0) = q^4
= (0.55)^4
= 0.55 * 0.55 * 0.55 * 0.55
= 0.09150625
≈ 9.15%
For 1 win, that would be the probability of 1 win by A and 3 wins by B.
You might think that would just be 0.45 * 0.55 * 0.55 * 0.55, but there are 4 ways to pick which of the matches is won by A.
P(x=1) = C(4,1) * p^1 * q^3
= 4 * (0.45) * (0.55)^3
= 4 * 0.45 * 0.55 * 0.55 * 0.55
= 0.299475
≈ 29.95%
P(x=0 or 1) = P(x=0) + P(x=1)
= 0.0915 + 0.2995
= 0.3910
= 39.1%
But you want the probability of 2 or more which is the opposite probability:
P(x ≥ 2) = 1 - P(x = 0 or 1)
= 1 - 0.3910
= 0.6090
= 60.9%
Answer:
P(x ≥ 2) = 60.9%