SOLUTION: A high school baseball player has a 0.296 batting average. In one game, he gets 5 at bats. What is the probability he will get at least 3 hits in the game?

Question 1161121: A high school baseball player has a 0.296 batting average. In one game, he gets 5 at bats. What is the probability he will get at least 3 hits in the game?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Let's compute the probability of getting exactly 3 hits

p = 0.296 is the probability of success = probability of getting a hit
k = number of hits we want = 3
n = sample size = number of at bats (aka batting attempts) = 5

Compute the binomial coefficient
Use the combination formula
n C k = (n!)/(k!*(n-k)!)
5 C 3 = (5!)/(3!*(5-3)!)
5 C 3 = (5!)/(3!*2!)
5 C 3 = (5*4*3!)/(3!*2!)
5 C 3 = (5*4)/(2!)
5 C 3 = (5*4)/(2*1)
5 C 3 = 20/2
5 C 3 = 10

This is then useful to compute the binomial probability
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 3) = (5 C 3)*(0.296)^(3)*(1-0.296)^(5-3)
P(X = 3) = (5 C 3)*(0.296)^(3)*(0.704)^(2)
P(X = 3) = (10)*(0.296)^(3)*(0.704)^2
P(X = 3) = (10)*(0.025934336)*(0.495616)
P(X = 3) = 0.12853471870976
This is the approximate probability of getting exactly 3 hits

We will use this value later. Let A = 0.12853471870976

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Keeping n = 5 and p = 0.296 the same, we will compute the probability when k = 4. This represents the probability of getting exactly 4 hits

n C k = (n!)/(k!*(n-k)!)
5 C 4 = (5!)/(4!*(5-4)!)
5 C 4 = (5!)/(4!*1!)
5 C 4 = (5*4!)/(4!*1!)
5 C 4 = (5)/(1!)
5 C 4 = (5)/(1)
5 C 4 = 5/1
5 C 4 = 5

P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 4) = (5 C 4)*(0.296)^(4)*(1-0.296)^(5-4)
P(X = 4) = (5 C 4)*(0.296)^(4)*(0.704)^(1)
P(X = 4) = (5)*(0.296)^(4)*(0.704)^1
P(X = 4) = (5)*(0.007676563456)*(0.704)
P(X = 4) = 0.02702150336512

We will use this value later. Let B = 0.02702150336512

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Repeat for k = 5. Keep n and p the same.

n C k = (n!)/(k!*(n-k)!)
5 C 5 = (5!)/(5!*(5-5)!)
5 C 5 = (5!)/(5!*0!)
5 C 5 = (1)/(1*0!)
5 C 5 = (1)/(1*1)
5 C 5 = (1)/(1)
5 C 5 = 1

P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 5) = (5 C 5)*(0.296)^(5)*(1-0.296)^(5-5)
P(X = 5) = (5 C 5)*(0.296)^(5)*(0.704)^(0)
P(X = 5) = (1)*(0.296)^(5)*(0.704)^0
P(X = 5) = (1)*(0.002272262782976)*(1)
P(X = 5) = 0.002272262782976

We will use this value later. Let C = 0.002272262782976

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We can now find the probability of getting at least 3 hits

Add up the values of A,B and C we found earlier
P(at least 3 hits) = P(exactly 3 hits)+P(exactly 4 hits)+P(exactly 5 hits)
P(at least 3 hits) = A + B + C
P(at least 3 hits) = 0.12853471870976 + 0.02702150336512 + 0.002272262782976
P(at least 3 hits) = 0.15782848485786

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The final answer is approximately 0.15782848485786
Round however you need to.

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