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%

Step-by-step 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

$P(A)=\frac{37}{56}$

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, $P(B\cap A)=\frac{24}{56}$

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

$P(B|A)=\dfrac{P(B\cap A)}{P(A)}\\\\=\dfrac{\dfrac{24}{56}}{\dfrac{37}{56}}\\\\\\=\dfrac{24}{37}=0.648648648649$

In percent, $P(B|A)= 0.648648648649\times100=64.8648648649\%\approx64.9\%$

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