STAT3000
Section 5.2:
The Binomial Distributions
Recall that about 45% of people have type O blood.
Suppose you select 3 people at random and check
their blood type.
What is the probability that all 3
have type O blood?
What is the probability that 2
have type O blood? What is the probability that only
1 has type O blood?
What is the probability that
none have type O blood?
S =
{OOO, OON,ONO, NOO, ONN, NON, NNO, NNN}
P(3 type O) = P(OOO) = (O.45) (O.45) (O.45)
= 1(0.45
3
) = 0.0911
P(2 type O) = P(OON or ONO or NOO)
= 3(0.45)(O.45)(0.55) = 3(0.45
2
)(0.55
1
)
= 0.3341
P(1 type O) = P(ONN or NON or NNO)
= 3(0.45)(0.55)(0.55) = 3(0.45
1
)(0.55
2
)
= 0.4084
P(no type O) = P(NNN)
=1 (0.55) (0.55) (0.55) = 1(0.55
3
)
= 0.1664
83
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View Full DocumentBinomial
Distribution
1.
There are a fixed number n of observations.
2.
The n observations are all independent
.
That is,
knowing the result of one observation tells you
nothing about the other observations.
3. Each observation falls into one of just two
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 Spring '08
 Staff
 Binomial, Normal Distribution, Probability, Probability theory, Binomial distribution, #, binomial random variable

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