E. Conditional Probabilities
A conditional probability is the probability of an event given
that another event has occurred.
Example 10
What
is the probability that the total of two dice will be greater
than 8 given that the first die is a 6?
Solution
This can be computed by considering only outcomes for which the
first die is a 6. Then, determine the proportion of these outcomes
that total more than 8. All the possible outcomes for two dice
are shown in the section on simple probability. There are 6 outcomes
for which the first die is a 6: (6,1),(6,2),(6,3),(6,4),(6,5),(6,6),
and of these, there are four that total more than 8. The probability
of a total greater than 8 given that the first die is 6 is therefore
4/6 = 2/3.
1. Probability of A and B
If A
and B are independent, then the probability that events A and
B both occur is p(A and B) = p(A) × p(B). In other words,
the probability of A and B both occurring is the product of the
probability of A and the probability of B. What is the probability
that a coin will come up with heads twice in a row? Two events
must occur: a heads on the first toss and a heads on the second
toss. Since the probability of each event is 1/2, the probability
of both events is: 1/2 × 1/2 = 1/4. Now consider a similar
problem: someone draws a card at random out of a deck, replaces
it, and then draws another card at random. What is the probability
that the first card is the ace of clubs and the second card is
a club (any club)?
Since there is only one ace of
clubs in the deck, the probability of the first event is 1/52.
Since 13/52 = 1/4 of the deck is composed of clubs, the probability
of the second event is 1/4. Therefore, the probability of both
events is 1/52 × 1/4 = 1/208.
What's the probability of A and
B (2 of 2) if A and B are not independent? If A and B are not
independent, then the probability of A and B is p(A and B) =
p(A) × p(B|A) where p(B|A) is the conditional probability
of B given A. If someone draws a card at random from a deck and
then, without replacing the first card, draws a second card,
what is the probability that both cards will be aces? Event A
is that the first card is an ace. Since 4 of the 52 cards are
aces, p(A) = 4/52 = 1/13. Given that the first card is an ace,
what is the probability that the second card will be an ace as
well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A)
= 3/51 = 1/17, and the probability of A and B is 1/13 × 1/17 = 1/221.
2. Probability of A or B
If
events A and B are mutually exclusive, then the probability of
A or B is simply:
p(A or B) = p(A) + p(B).
What is the probability of rolling
a die and getting either a 1 or a 6? Since it is impossible
to get both a 1 and a 6, these two events are mutually exclusive.
Therefore,
p(1 or 6) = p(1) + p(6) = 1/6 + 1/6 = 1/3
If the events A and B are not mutually exclusive, then
p(A or B) = p(A) + p(B) - p(A and B).
The logic behind this formula is that when p(A) and p(B) are
added, the occasions on
which A and B both occur are counted twice. To adjust for this,
p(A and B) is subtracted.
Example 11
What is the probability that a card selected from a deck will
be either an ace or a spade?
Solution
The relevant probabilities are
p(ace) = 4/52
p(spade) = 13/52
The only way an ace and a spade
can both be drawn is to draw the ace of spades. There is only
one ace of spades, so p(ace and spade) = 1/52. The probability
of an ace or a spade can be computed as p(ace)+p(spade)-p(ace
and spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.
w
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w F. Combinations
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