Section 4.6 Probability
307
Definition CONditiONal PROBaBility
Given events E 1 and E 2 , the conditional probability of E 2 given E 1 , P(E 2 0 E 1 ), is
P(E 2 0 E 1 ) =
P(E 1 d E 2 )
P(E 1 )
PraCtiCe 45 In the problem of tossing a fair coin twice, what is the probability of getting two heads
given that at least one of the tosses results in heads? (Hint: Let E 2 be the event of two heads and E 1 be
the event of at least one head.)
■
example 71
In a drug study of a group of patients, 17% responded positively to compound A,
34% responded positively to compound B, and 8% responded positively to both.
The probability that a patient responded positively to compound B given that he or
she responded positively to A is
P(B 0 A) =
P(A d B)
P(A)
=
0.08
0.17
> 0.47
If P(E 2 0 E 1 ) = P(E 2 ), then E 2 is just as likely to happen whether E 1 happens or
not. In this case E 1 and E 2 are said to be independent events and we have
P(E 2 0 E 1 ) =
P(E 1 d E 2 )
P(E 1 )
= P(E 2 )
or
P(E 1 d E 2 ) = P(E 1 ) # P(E 2 )
(2)
Equation (2) can be extended to any finite number of independent events and can
also be used to test whether events are independent.
example 72
The events of tossing a coin and coming up heads one time (E 1 ) and heads the next
(E 2 ) are independent events because
P(E 1 d E 2 ) = 1/4
P(E 1 ) = 1/2, P(E 2 ) = 1/2
so Equation (2) is satisfied. If we toss a fair coin repeatedly and the coin lands 4 or
5 or 6 times in a row heads up, we may feel that tails is “due to come up,” in other
words, that at the next toss the coin has a better than 50% probability of coming up
tails, but in fact that’s not the case. It is true that the probability of getting longer
and longer runs of heads decreases from 1/4 (two heads) to 1/8 (three heads) to
1/16 (four heads), and so forth, yet on each successive toss, the probability of getting a head is still 1/2.
307
Definition CONditiONal PROBaBility
Given events E 1 and E 2 , the conditional probability of E 2 given E 1 , P(E 2 0 E 1 ), is
P(E 2 0 E 1 ) =
P(E 1 d E 2 )
P(E 1 )
PraCtiCe 45 In the problem of tossing a fair coin twice, what is the probability of getting two heads
given that at least one of the tosses results in heads? (Hint: Let E 2 be the event of two heads and E 1 be
the event of at least one head.)
■
example 71
In a drug study of a group of patients, 17% responded positively to compound A,
34% responded positively to compound B, and 8% responded positively to both.
The probability that a patient responded positively to compound B given that he or
she responded positively to A is
P(B 0 A) =
P(A d B)
P(A)
=
0.08
0.17
> 0.47
If P(E 2 0 E 1 ) = P(E 2 ), then E 2 is just as likely to happen whether E 1 happens or
not. In this case E 1 and E 2 are said to be independent events and we have
P(E 2 0 E 1 ) =
P(E 1 d E 2 )
P(E 1 )
= P(E 2 )
or
P(E 1 d E 2 ) = P(E 1 ) # P(E 2 )
(2)
Equation (2) can be extended to any finite number of independent events and can
also be used to test whether events are independent.
example 72
The events of tossing a coin and coming up heads one time (E 1 ) and heads the next
(E 2 ) are independent events because
P(E 1 d E 2 ) = 1/4
P(E 1 ) = 1/2, P(E 2 ) = 1/2
so Equation (2) is satisfied. If we toss a fair coin repeatedly and the coin lands 4 or
5 or 6 times in a row heads up, we may feel that tails is “due to come up,” in other
words, that at the next toss the coin has a better than 50% probability of coming up
tails, but in fact that’s not the case. It is true that the probability of getting longer
and longer runs of heads decreases from 1/4 (two heads) to 1/8 (three heads) to
1/16 (four heads), and so forth, yet on each successive toss, the probability of getting a head is still 1/2.