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About Paul Klarreich
Expertise
I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction.
I can also try (but not guarantee) to answer questions on Abstract Algebra
-- groups, rings, etc. and Analysis -- sequences, limits, continuity.
I won't understand specialized engineering or business jargon.
Experience
I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.
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You are here: Experts > Science > Mathematics > Advanced Math > Binomial expansion.
Expert: Paul Klarreich
Date: 5/16/2008
Subject: Binomial expansion.
Question
When Susan and Jessica play a card game, Susan wins 60% of the time. If they play 9 games, what is the probability that Jessica will have won more games than Susan?
Answer
Questioner: Wendy
Category: Advanced Math
Private: No
Subject: Probability
Question: When Susan and Jessica play a card game, Susan wins 60% of the time. If they play 9 games, what is the probability that Jessica will have won more games than Susan?
...............................
Hi, Wendy,
In general, if w = p(Wendy wins a game), and
j = (1 - w) = p(Jessica), and n games are played, we have:
p(Wendy wins exactly k games) = C(n,k) w^k (1- w)^(n-k)
Naturally,
SUM[k=0 to n] C(n,k) w^k (1- w)^(n-k) = (w + (1-w))^n = 1
In this case, n = 9, and you want Wendy to win 5 to 9 games:
SUM[k=5 to 9] C(9,k) w^k (1- w)^(9-k)
C(9,5) w^5 (1- w)^(4)
C(9,6) w^6 (1- w)^(3)
C(9,7) w^7 (1- w)^(2)
C(9,8) w^8 (1- w)^(1)
C(9,9) w^9 (1- w)^(0)
And, since w = 3/5, 1-w = 2/5, that comes to
C(9,5)(3/5)^5 (2/5)^4 + C(9,6)(3/5)^6 (2/5)^3 + C(9,7)(3/5)^7 (2/5)^2 + C(9,8)(3/5)^8 (2/5)^1 + C(9,9)(3/5)^9 (2/5)^0 =
[ C(9,5)(3)^5 (2)^4 + C(9,6)(3)^6 (2)^3 + C(9,7)(3)^7 (2)^2 + C(9,8)(3)^8 (2)^1 + C(9,9)(3)^9 (2)^0] over 5^9
which you can work out now.
Excel is a good tool for the calculation. Mine looks like this:
k C(9,k) 3^k*2^(9-k) p[k]
5 126 3888 0.250822656
6 84 5832 0.250822656
7 36 8748 0.161243136
8 9 13122 0.060466176
9 1 19683 0.010077696
Total = 0.73343232, which is your answer. Interesting that the first two entries in the p[k] column are identical. See if you can prove that.
BTW, look for a recent answer in this category.
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