Question

The table shows whether members in a fishing group caught fish or did not catch fish and whether they fished from a boat or from the shore. One member is randomly selected.

What is the probability that the members fished from a boat, given that he or she caught fish?
Write the probability as a percent. Round to the nearest tenth of a percent as needed.

Answer 1

Answer: 64.9%

Explanation:

From the given table , the number of members caught fish =
`24+13=37`

Total Members =
`24+11+13+8=56`

Let A be the event that members caught fish , then

`P(A)=(37)/(56)`

Let B be the event of members fished from a boat.

The number of members fished from boat and caught fish =
`B\cap A=24`

Then,
`P(B\cap A)=(24)/(56)`

Now, the probability that the members fished from a boat, given that he or she caught fish is given by :-

`P(B|A)=(P(B\cap A))/(P(A))\\\\=((24)/(56))/((37)/(56))\\\\\\=(24)/(37)=0.648648648649`

In percent,
`P(B|A)= 0.648648648649*100=64.8648648649\%\approx64.9\%`

Hence, the probability that the members fished from a boat, given that he or she caught fish=64.9%

Answer 2

The given scenario falls under the category of conditional probability. There are a total of 56 people in the fishing group described above. The probability that the a person caught a fish and fished from a boat is 24/56. The probability that he fished from a boat is (24+11)/56. For the conditional probability in this item is,
P = (24/56) / ((24+11)/56) = 24/35 = 0.69