Question 1 of 5
P(H and W)
P(W | H)
P(H | W)
P(H)
P(W)
Good job! We are limiting our interest to the home games only, so we
condition on H. Out of the home games, we are interested in the probability of
winning (W), therefore we are interested in P(W | H).
Question 2 of 5
.15
The first three questions refer to the following information:
Suppose a basketball team had a season of games with the following characteristics:
•
60% of all the games were at-home games. Denote this by H (the
remaining were away games).
•
35% of all games were wins. Denote this by W (the remaining
were losses).
•
25% of all games were at-home wins.
Of the at-home games, we are interested in finding what proportion were wins.
In order to figure this out, we need to find:
Again here is the information about the characteristics of a basketball team's
season:
•
60% of all the games were at-home games. Denote this by H (the
remaining were away games).
•
35% of all games were wins. Denote this by W (the remaining
were losses).
•
25% of all games were at-home wins.
Of the at-home games, what proportion of games were wins? (Note: Some
answers are rounded to two decimal places.)
Question 3 of 5
.09
.15
.21
.42
.71
Good job! Given that the team won the game (W), how likely is it that this was
a home game (H)? Therefore, we are interested in P(H | W). Again, using the
definition of conditional probability:
Question 4 of 5
Again here is the information about the characteristics of a basketball team's
season:
•
60% of all the games were at-home games. Denote this by H (the
remaining were away games).
•
35% of all games were wins. Denote this by W (the remaining
were losses).
•
25% of all games were at-home wins.
If the team won a game, how likely is it that this was a home game? (Note:
Some answers are rounded to 2 decimal places.)
.21
.25
.42
.71
Good job! We want to find P(W | H). We apply the definition of conditional
probability:
Cannot find it since P(B) is not known.
Cannot find it since P(A and B) is not known.
Cannot find it since both P(B) and P(A and B) are not known.
It is equal to .5.
It is equal to .25.
Good job! If two events are independent, then P(A | B) = P(A) [knowing that B
occurs has no impact on the probability that A occurs]. Therefore, if we are
given P(A) = .5, and that A and B are independent, then it must also be true
that P(A | B) = .5.
No, since .31 * .38 is not equal to .42.
No, since .31 is not equal to .42.
No, since .38 is not equal to .42.
Let A and B be two independent events. If P(A) = .5, what can you say about
P(A | B)?
Question 5 of 5
Dogs are inbred for such desirable characteristics as blue eye color; but an
unfortunate by-product of such inbreeding can be the emergence of
characteristics such as deafness. A 1992 study of Dalmatians (by Strain and
others, as reported in The Dalmatians Dilemma) found the following:
Based on the results of this study is "having blue eyes" independent of "being
deaf"?
(i)
31% of all Dalmatians have blue eyes.
(ii)
38% of all Dalmatians are deaf.
(iii)
42%
of blue-eyed Dalmatians are deaf.
Yes, since .31 * .38 is not equal to .42.
Yes, since .38 is not equal to .42.
This is not quite right. Knowing that P(blue eyes) ≠ P (deaf | blue eyes) doesn't
tell us anything useful. Recall that having blue eyes and being deaf are
independent if knowing a Dalmatian has blue eyes doesn't change the
probability that it is deaf. This would mean that P(deaf | blue eyes) = P(deaf).
Consider the remaining options. (B) is the correct answer.
