Math331, Fall 2008
Instructor: David Anderson
Section 3.4 Homework Answers
Homework: pgs. 105 - 106 #’s 1, 2, 8.
1. We consider one transmission. We let Abe the event that a dot was transmitted. We let
Bbe the event that a dot was received. Note: the sample space here is S={(a, b)|a, b ∈
{dot, dash}}, where aand brepresent what was transmitted and received, respectively. We
want P(A|B). Bayes’ formula gives us the answer:
P(A|B) = P(B|A)P(A)
P(B|A)P(A) + P(B|Ac)P(Ac)=.75 ∗.4
.75 ∗.4 + (1/3) ∗.6=.6.
2. Let Abe the event that a question was marked correctly. Let Bbe the event that the
student actually knew the answer and did not guess. Then we want P(B|A). We again use
Bayes’ formula
P(B|A) = P(A|B)P(B)
P(A|B)P(B) + P(A|Bc)P(Bc)=1∗(2/3)
1∗(2/3) + (1/4) ∗(1/3) =8
9.
8. Let Bbe the event that the coin selected from urn I is a dime. Let Abe the event that
two of the three coins are dimes. Then,
P(B|A) = P(A|B)P(B)
P(A|B)P(B) + P(A|Bc)P(Bc).
It is easy to see P(B) = 4/7, and P(Bc) = 3/7. Therefore,
P(B|A) = P(A|B)(4/7)
P(A|B)(4/7) + P(A|Bc)(3/7) =4P(A|B)
4P(A|B) + 3P(A|Bc).
To find P(A|B) we realize that to get two dimes total, if one dime is from urn I, then only
one can come from the other two urns. Therefore, we need to find the probability of selecting
only one dime from the two remaining urns. This is given by:
P(A|B) = (2/7)(1/4) + (5/7)(3/4) = 17/28.
Similarly, we see P(A|Bc) as the probability that we chose dimes from both of the other
urns. Therefore
P(A|B) = (5/7)(1/4) = 5/28.
Therefore,
P(B|A) = 4P(A|B)
4P(A|B) + 3P(A|Bc)
=4∗(17/28)
4∗(17/28) + 3 ∗(5/28) =4∗17
4∗17 + 3 ∗5=68
83 = 0.819277.
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