When Peter plays Paul, Peter wins on avg 2 games out of 3 played. If Peter and Paul play 6 giames, what is the

Question

probability that

peter wins exactly 4 of the games

Paul wins more games than Peter


So how do I incorporate the binomial distribution into this?

Answer 1

Let
p = probability Peter wins a game
q = probability Peter loses a game

p = 2/3
q = 1 - 2/3 = 1/3

The expansion of (p + q)^6 is
1 6 15 20 15 6 1

Let P(4) = P(Peter wins Exactly 4 of 6)

P(4) = 15(p^4)q² = 15[(2/3)^4] [(1/3)²]
P(4) = 15*2^4 / 3^6 = 240/729
P(4) = 80/243 ≈ .329
___________

Let P(x) = P(Peter wins Exactly x of 6)
Let P(Paul) = P(Paul wins more than Peter)

P(Paul) = P(0) + P(1) + P(2)
P(Paul) = (p^0)(q^6) + 6(p)(q^5) + 15(p²)(q^4)
P(Paul) = (1/3)^6 + 6(2/3) [(1/3)^5] + 15[(2/3)²] [(1/3)^4]
P(Paul) = (1 + 6*2 + 15*2²) / 3^6 = 73/729 ≈ 0.100

Answer 2

2