Binominal Distributions Worksheet With Answer Key - Helm 2008 Section 372 Page 5
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n n n = = = 2 2 2
Here there are 3 possibilities: We can observe 2, 1 or 0 successes. Let S denote a success
and F denote a failure. So a failure followed by a success would be denoted by F S whilst two failures
followed by one success would be denoted by F F S and so on.
Then
2
P(2 successes in 2 trials) = P(SS) = P(S)P(S) = p
(where we have used the assumption of independence between trials and hence multiplied probabili-
ties). Now, using the usual rules of basic probability, we have:
P(1 success in 2 trials) = P[(SF )
(F S)] = P(SF ) + P(F S) = pq + qp = 2pq
2
P(0 successes in 2 trials) = P(F F ) = P(F )P(F ) = q
2
2
The three probabilities we have found
q
, 2qp, p
are in fact the terms which arise in the
2
2
2
binomial expansion of (q + p)
= q
+ 2qp + p
. We also note that since q = 1
p the probabilities
sum to 1 (as we should expect):
2
2
2
2
q
+ 2qp + p
= (q + p)
= ((1
p) + p)
= 1
Task
List the outcomes for the binomial model for the case n = 3, calculate their
probabilities and display the results in a table.
Your solution
Answer
three successes, two successes, one success, no successes
3
Three successes occur only as SSS with probability p
.
2
Two successes can occur as SSF with probability (p
q), as SF S with probability (pqp) or as F SS
2
with probability (qp
).
2
These are mutually exclusive events so the combined probability is the sum 3p
q.
Similarly, we can calculate the other probabilities and obtain the following table of results.
Number of successes
3
2
1
0
3
2
2
3
Probability
p
3p
q 3pq
q
21
HELM (2008):
Section 37.2: The Binomial Distribution
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