Question

A population of 80 rats is tested for 4 genetic mutations after exposure to some chemicals: mutation A, mutation B, mutation C, and mutation D. 43 rats tested positive for mutation A. 37 rats tested positive for mutation B. 39 rats tested positive for mutation C. 35 rats tested positive for mutation D. One rat tested positive for all four mutations, 5 rats tested positive for mutations A, B, and C. 4 rats tested positive for mutations A, B, and D. 6 rats tested positive for mutations A, C, and D. 3 rats tested positive for mutations B, Cand D. 64 rats tested positive for mutations A or B. 63 rats tested positive for mutations A or C.59 rats tested positive for mutations A or D. 58 rats tested positive for mutations B or C. 59 rats tested positive for mutations B or D. 60 tested positive for mutations Cor D. 8 rats did not show any evidence of genetic mutation What is the probability that if 5 rats are selected at random, 3 will have exactly 2 genetic mutations? Round your answer to five decimal places.

Answer 1

To find the probability, use the combination formula to calculate the number of ways to choose 3 rats with exactly 2 mutations and 2 rats with no mutations, and divide it by the total number of ways to choose any 5 rats.

Step-by-step explanation:

To find the probability that 3 out of 5 rats will have exactly 2 genetic mutations, we need to calculate the total number of ways we can choose 3 rats to have 2 mutations and 2 rats to have no mutations, divided by the total number of ways to choose any 5 rats. Let's calculate it step by step:

1. There are 80 rats in total and one rat tested positive for all four mutations. So, there are 79 rats remaining.
2. Since 5 rats are selected at random, the total number of ways to choose any 5 rats is given by the combination formula: C(79, 5).
3. Next, we calculate the number of ways to choose 3 rats with exactly 2 mutations and 2 rats with no mutations. There are 4 mutations in total and we need to choose 2 of them for the 3 selected rats. The remaining 2 rats should have no mutations. So, the number of ways to choose the 3 rats is given by: C(4, 2) * C(75, 2).
4. Finally, we divide the number of ways to choose 3 rats with exactly 2 mutations and 2 rats with no mutations by the total number of ways to choose any 5 rats to get the probability.

Answer 2

To solve this problem, we need to use the concept of hypergeometric distribution, which gives the probability of selecting a certain number of objects with a specific characteristic from a population of known size without replacement. We will use the formula:

P(X = k) = [ C(M, k) * C(N - M, n - k) ] / C(N, n)

where:

P(X = k) is the probability of selecting k objects with the desired characteristic;

C(M, k) is the number of ways to select k objects with the desired characteristic from a population of M objects;

C(N - M, n - k) is the number of ways to select n - k objects without the desired characteristic from a population of N - M objects;

C(N, n) is the total number of ways to select n objects from a population of N objects.

In our case, we want to select 5 rats out of a population of 80, and we want exactly 3 of them to have 2 genetic mutations. We can calculate this probability as follows:

P(3 rats have exactly 2 mutations) = [ C(12, 3) * C(68, 2) ] / C(80, 5)

where:

M is the number of rats that have exactly 2 mutations, which is the sum of the rats that have mutations AB, AC, AD, BC, BD, and CD, or M = 5 + 6 + 4 + 3 + 3 + 1 = 22;

N - M is the number of rats that do not have exactly 2 mutations, which is the remaining population of 80 - 22 = 58 rats;

n is the number of rats we want to select, which is 5.

We can simplify this expression as follows:

P(3 rats have exactly 2 mutations) = [ C(12, 3) * C(68, 2) ] / C(80, 5)

= [ (12! / (3! * 9!)) * (68! / (2! * 66!)) ] / (80! / (5! * 75!))

= 0.03617

Therefore, the probability that if 5 rats are selected at random, 3 will have exactly 2 genetic mutations is 0.03617 (rounded to five decimal places).