Question

Ten offspring are generated from a test cross between a Cc heterozygote and a cc homozygote. What is the probability that exactly nine of the offspring have the cc genotype? 0.0098 0.064 0.05 0.00098 0.0064

Answer 1

The probability of having exactly nine offspring with the cc genotype out of ten offspring is 0.064.

How to find probability?

The probability of having exactly nine offspring with the cc genotype out of ten offspring can be calculated using the binomial probability formula:

`P(k) = (n_(choose) k) * p^k * (1-p)^{(n-k)`

where:

P(k) = probability of having exactly k successes in n trials

n = number of trials (in this case, 10 offspring)

k = number of successes (in this case, 9 offspring with the cc genotype)

p = probability of success (in this case, the probability of having an offspring with the cc genotype)

(1-p) = probability of failure (in this case, the probability of having an offspring with the Cc genotype)

The probability of having an offspring with the cc genotype is 1/2, since one parent is cc and one parent is Cc. Therefore,

p = 1/2 and (1-p) = 1/2.

Plugging in the values:

P(9) = (10 choose 9) × (1/2)⁹ × (1/2)¹

P(9) = 125970 × 1/512 × 1/2

P(9) ≈ 0.064

Therefore, the probability of having exactly nine offspring with the cc genotype out of ten offspring is approximately 0.064.