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Answer :
The owner needs to inseminate at least 6 cows to be 90% confident that the bull is not a carrier of the recessive lethal condition.
To calculate the number of cows the owner needs to inseminate in order to be 90% confident that the bull is not a carrier of the recessive lethal condition, we can use the concept of a binomial distribution.
Let's denote:
p = probability of a cow being a carrier (heterozygous) = 0.65
q = probability of a cow not being a carrier (homozygous dominant) = 0.35
n = total number of cows inseminated
k = number of cows that need to be inseminated to be 90% confident
We want to find the smallest value of k for which the cumulative probability of getting at least one carrier among the inseminated cows is greater than or equal to 0.90.
The cumulative probability of getting at least one carrier can be calculated as:
P(X ≥ 1) = 1 - P(X = 0)
Using the binomial distribution formula, we have:
[tex]P(X = k) = C(n, k) \times p^k \times q^{(n-k)[/tex]
Setting up the equation:
[tex]1 - P(X = 0) = 1 - C(n, 0) \times p^0 \times q^n[/tex] ≥ 0.90
Simplifying:
[tex]1 - q^n[/tex] ≥ 0.90
Now we can solve for n using logarithms:
[tex]q^n[/tex] ≤ 0.10
n [tex]\times[/tex]log(q) ≤ log(0.10)
n ≥ log(0.10) / log(q)
Using the given values:
n ≥ log(0.10) / log(0.35)
Using a calculator:
n ≥ 5.572
Since we cannot have a fraction of a cow, the owner needs to inseminate at least 6 cows to be 90% confident that the bull is not a carrier of the recessive lethal condition.
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