let's look at each possibility in turn.
first hit only:
the probability that the group gets a hit on the first record is .28.
assuming this happens, then the probability that the group gets a hit on the second and third record is 1.0 each because the second and third records are automatically hits if the first record is a hit, as given by the problem statement.
what this says is that the first record can't be a hit if the group can only have 1 hit out of 3 records, because once the group has a hit on their first record, the following 2 records are automatically hits, which means their probability of failure is 0.
therefore the first record can't be a hit, so the probability that the first record only is a hit is therefore 0.
the probability that the first record is not a hit is 1 - .28 = .72.
given that the first record is not a hit, then the probability of the second record being a hit is .14 and the probability of the third record being a hit is also .14.
the probability that the second record only is a hit is therefore:
.72 * .14 * .86 = .086688.
.72 is the probability that the first record is not a hit.
.14 is the probability that the second record is a hit given that the first record is not a hit.
.86 is the probability that the third record is not a hit given that the first record is not a hit and the second record is a hit.
given that the first record is not a hit and the second record is not a hit, then the probability that the third record is a hit is .07.
the probability that the third record only is a hit is therefore:
.72 * .86 * .07 = .043344.
.72 is the probability that the first record is not a hit.
.86 is the probability that the second record is not a hit given that the first record is not a hit.
.07 is the probability that the third record is a hit given that the first record is not a hit and the second record is not a hit.
the total probability that only 1 record out of 3 is a hit would therefore be 0 + .086688 + .043344 = .130032.
the following tree diagrams show that the total probability is 1 as it should be.
there are 3 events with 2 possible outcomes in each event.
therefore the total number of possible events is 2^3 = 8.
the 3 events are the number of records that are published.
the possible outcomes for each event are h or n.
h is the probability that the record is a hit.
n is the probability that the record is not a hit.
i was hoping that the total probability would be 1 because that's what it should be if the problem is well designed.
kudos to the people who designed the problem, because the total probability was equal to 1, as it should have been.
my tree diagram and associated calculations are shown below:
the total probability that only record is a hit out of 3 tries is options d, f, and g.
p(d) is 0
p(f) is .086688
p(g) is .043344.
total of p(d) + p(f) + p(g) is .130032 as calculated earlier.
