- Prob(S) = Prob(Success) is the same in each trial.
- The random variable X counts the number of successes.
n = number of times a trial is repeated
p = P(S) = Prob( a success in a single trial)
q = P(F) = Prob( a failure in a single trial)
X = the random variable
Binomial Probability Formula :
Prob(X out of n)=
There are many conditions or assumptions necessary before we proceed to the
calculations of this particular distribution.
A binomial experiment is a probability experiment that satisfies the following conditions:
- The experiment is repeated for a fixed number of times (n) where each trial
is independent of the others.
- Only TWO possible outcomes are possible for each trial. Usually we describe
the outcomes as Success(S) or Failure(F).
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Example question :
Multiple
choice tests
George has no back- ground in the test he is taking and is purely guessing
on his multiple-choice test. There are 10 questions and each question has
four possible answers.
What are his chances of getting 8 out of 10 correct or better?
- Show that this is a binomial probability distribution.
- Find the probabilities for getting better than 8 out of 10 correct.
- As far as we can tell George is faced with 10 separate trials.(n=10)
The trials are independent.
There only two possible outcomes for each question:
correct(Success) or incorrect (Failure).
Prob(S) for each trial is 1/4
Prob(F) for each trial is 3/4
X is the random variable for number of successes.
Yes it is a binomial distribution.
- We must calculate the probability of his getting 8 of 10 correct and 9
out of 10 correct and getting 10 out of 10 correct.
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then add the three results: Binomial Probability Distributions may look graphically as follows :
Example 1
Probability(George
gets 1 or 2 or 3 or 4 correct) is much larger than Prob(8 or 9 or 10) correct
answers
| X | P(X) | |
| Prob( George gets 0 on 10) | 0 | 0.0563 |
| Prob(George gets 1 on 10) | 1 | 0.1877 |
| Prob(George gets 2 on 10) | 2 | 0.2816 |
| Prob(George gets 3 on 10) | 3 | 0.2503 |
| Prob(George gets 4 on 10) | 4 | 0.1460 |
| Prob(George gets 5 on 10) | 5 | 0.0584 |
| Prob(George gets 6 on 10) | 6 | 0.0162 |
| Prob(George gets 7 on 10) | 7 | 0.0031 |
| Prob(George gets 8 on 10) | 8 | 0.0004 |
| Prob(George gets 9 on 10) | 9 | 0.000029 |
| Prob(George gets 10 on 10) | 10 | 0.000001 |
| Prob( 8 or 9 or 10 ) is equal to 0.0004 or 0.04% | 0.0004 | |
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In Canada Prob(a boy) = 0.52 and Prob(a girl) = 0.48
Based on these probabilities analyze the following 4-child families:
a) How many different-gender families are possible ? (# of girls + # of boys)
b) What are probabilities of each family of 4 children?
Answer
a) |
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Answer
b) We'll use Binomial calculations and call
Prob( a girl ) = P(S) = 0.48
Therefore P(F) = 0.52
X = number of girls of of 4. We have X = 4 , 3 , 2 , 1 , 0.
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Question:
We're
told that 38% of Canadians have type O+ blood.
Only two randomly selected people are chosen.
Give
the probabilities of selecting 0, 1 and 2 type O+ individuals.
Answer:
Prob( 0 type O+ out of 2 ) = 0.3844 or 38.4%
Prob( 1 type O+ out of 2 ) = 0.4712 or 47.1%
Prob( 2 type O+ out of 2 ) = 0.1444 or 14.4%
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