If a baseball player's batting average is 0.260 (26%), find the probability that the player will get at most 21 hits in 100 times at bat.

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- Feb 18th 2007, 09:19 AMgracyprobablity with normal distribution
If a baseball player's batting average is 0.260 (26%), find the probability that the player will get at most 21 hits in 100 times at bat.

- Feb 18th 2007, 12:39 PMSoroban
Hello, Gracy!

Who assigned this problem? .Professor deSade?

Quote:

If a baseball player's batting average is 0.260 (26%),

find the probability that the player will get at most 21 hits in 100 times at bat.

Calculate the probabilities of 0 hits, 1 hit, 2 hits, 3 hits, . . . 21 hits

. . and add them.

Here's the list to add . . .

0 hits: . C(100,0)·0.26)^0·(0.74)^100

1 hit: . .C(100,1)·(0.26)^1·(0.74)^99

2 hits: . C(100,2)·(0.26)^2·(0.74)^98

3 hits: . C(100,3)·(0.26)^3·(0.74)^97

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

21 hits: .C(100,21)·(0.26)^21·(0.74)^79

I'll wait in the car . . . - Feb 18th 2007, 01:23 PMThePerfectHacker
I would have done the same solution as you. But I believe there is an approximation, called the "Normal Distribution". I know that there is a much faster way of doing this, and I am sure Captain

**Blank[**will do that sometime later. I am just not sure how it works, probability is not where I am good.