Exposure of certain fruit fly population to an insecticide has produce

Michelle Isakson

Michelle Isakson

Answered question

2021-11-20

Exposure of certain fruit fly population to an insecticide has produced 2 kinds of mutations. 20% have wing mutation. 15% have eye mutation and 5% have both. A fly is selected at random.
a) If it has wing mutation, what is the probability that, it also has the eye mutation?
b) If it has the eye mutation, what is the probability that it has also the wing mutation?
c) What is the probability that it has at least one of the mutations?

Answer & Explanation

giskacu

giskacu

Beginner2021-11-21Added 22 answers

Step 1
Given Information:
20% have wing mutation i.e., P(W)=0.20
15% have eye mutation i.e., P(E)=0.15
5% have both i.e., P(W and E)=P(E and W)=0.05
A fly is selected at random.
(a) If it has the wing mutation, to find the probability that it also has the eye mutation:
Conditional probability is given by the formula:
P(AB)=P(A and B)P(B)
Required probability is obtained as follows:
P(EW)=P(E and W)P(W)
=0.050.20
=0.25
Thus, if it has the wing mutation, to find the probability that it also has the eye mutation is 0.25.
Step 2
(b) If it has the eye mutation, to find the probability that it also has the wing mutation:
Using the conditional probability formula, required probability is obtained as follows:
P(WE)=P(W and E)P(E)
=0.050.15
=0.3333333
0.33
Thus, if it has the eye mutation, to find the probability that it also has the wing mutation is 0.33
(c) To find the probability that it has at least one of the mutations:
Required probability:
P(at least one mutation)=1P(none of the mutations)
=1P(no wing mutationno eye mutation)
=1[(10.20)(10.15)]
=1[0.80.85]
=10.68
=0.32
Thus, the probability that it has at least one of the mutations is 0.32.

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