Math 10B Spring 2018 Discrete Probability Worksheet 3 Solutions
Discrete Probability Worksheet 3 Solutions
1. Suppose you flip two fair coins. Let Abe the event that the first coin is heads, Bthe event
that the second coin is heads and Cthe event that both coins show the same face. Are Aand
Bindependent? Aand C?Band C? How about A,B, and C?
Solution: At first, it might seem that only Aand Bare independent. But often our
intuition about probability can be misleading. So let’s try directly checking which are
independent using the definition of independence.
Let’s define our sample space to be all pairs of tails and heads. So there are 4 outcomes
and since the coins are fair, each outcome is equally likely. Now observe that each of the
events A,B, and Cconsists of two outcomes (the two outcomes for Care HH and T T ).
So each of these events has probability 1/2. In other words,
P(A) = P(B) = P(C) = 1
2.
The event A∩Bjust means that both coins show heads. The event A∩Cmeans that
the first coin is heads and both coins are the same—in other words, A∩Cjust means
that both coins are heads. And so does the event B∩C. So A∩B,A∩Cand B∩Care
actually all the same event, and all have the same probability, 1/4. In other words,
P(A∩B) = P(A∩B) = P(B∩C) = 1
4.
The event A∩B∩Cmeans that the first coin is heads, the second coin is heads, and both
coins are the same. But this, too, is just equivalent to saying that both coins are heads.
So we also have
P(A∩B∩C) = 1
4.
Using the definition of independence, all of this means that the following are independent:
Aand B
Aand C
Band C
but A,B, and Care not independent.
2. Show that if A,B, and Care independent events then
P(A|B∩C) = P(A)
Solution: Using the definitions of conditional probability and independence,
P(A|B∩C) = P(A∩(B∩C))
P(B∩C)=P(A∩B∩C)
P(B∩C)=P(A)P(B)P(C)
P(B)P(C)=P(A).
