SOLUTION: A high school baseball player has a 0.153 batting average. In one game, he gets 6 at bats. What is the probability he will get at least 2 hits in the game?

Question 1201351: A high school baseball player has a 0.153 batting average. In one game, he gets 6 at bats. What is the probability he will get at least 2 hits in the game?
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Answer: 0.230583
This value is approximate.

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Work Shown:

n = 6 = sample size = number of at bats
p = 0.153 = batting average = probability of success of any at bat

x = number of successful attempts = number of hits
x is some integer in the set {0, 1, 2, 3, 4, 5, 6}
x is between 0 and 6, including each endpoint.

B(x) = binomial probability
B(x) = (n C x)*(p^x)*(1-p)^(n-x)
B(x) = (6 C x)*(0.153^x)*(1-0.153)^(6-x)
B(x) = (6 C x)*(0.153^x)*(0.847)^(6-x)
The nCx refers to the nCr combination formula.

"at least 2 hits" translates to "2 or more hits"
We want to find the probability that x = 2 or x > 2
We need to calculate B(2)+B(3)+B(4)+B(5)+B(6)
A shortcut is to calculate B(0)+B(1) first and then subtract from 1.

Plug in x = 0
B(x) = (6 C x)*(0.153^x)*(0.847)^(6-x)
B(0) = (6 C 0)*(0.153^0)*(0.847)^(6-0)
B(0) = 1*(0.153^0)*(0.847)^6
B(0) = 0.36923296009284
B(0) = 0.369233

Plug in x = 1
B(x) = (6 C x)*(0.153^x)*(0.847)^(6-x)
B(1) = (6 C 1)*(0.153^1)*(0.847)^(6-1)
B(1) = 6*(0.153^1)*(0.847)^5
B(1) = 0.40018401105694
B(1) = 0.400184

Add the results
B(0)+B(1) = 0.369233 + 0.400184 = 0.769417
The probability of at most 1 hit is roughly 0.769417

The complement of this is
1-0.769417 = 0.230583 which is the approximate final answer.
The probability of at least 2 hits is roughly 0.230583 (i.e. about a 23.0583% chance).

Why did I subtract from 1?
Well consider the fact that
B(0) + B(1) + B(2) + B(3) + B(4) + B(5) + B(6) = 1
All of the probability values of a distribution must add to 1.

We already calculated the stuff highlighted in blue
B(0) + B(1) + B(2) + B(3) + B(4) + B(5) + B(6) = 1
and the stuff in red represents what we want to calculate.

We can subtract the blue stuff from both sides to isolate the red stuff.
B(2) + B(3) + B(4) + B(5) + B(6) = 1 - ( B(0) + B(1) )

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Here are a few binomial distribution calculators.
https://www.gigacalculator.com/calculators/binomial-probability-calculator.php
https://www.omnicalculator.com/statistics/binomial-distribution
You could also use a spreadsheet or a TI83/TI84 calculator.

Here is an article talking about binomial probabilities using a TI84
https://www.statology.org/binomial-probabilities-ti-84-calculator/
For this particular problem, we can input 1 - binomcdf(6, 0.153, 1) into the TI84.
This will add up the values from B(0) to B(1).
Then subtract the result from 1 to get the final answer.
Be sure to use the CDF and not the PDF.
The PDF is one specific value, while the CDF adds up multiple values below a specific x value.

Another question involving the binomial probability distribution
https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1201223.html

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