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)   (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%

 

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