Answer: Probability that two people have the same number of repeats in both location A and location B is [tex]\dfrac{1}{40000
Step-by-step explanation:
Since we have given that
Probability that two people both have 3 repeat sequences at location A = [tex]\dfrac{1}{500}[/tex]
Probability that two people both have 4 repeat sequences at location B = [tex]\dfrac{1}{800}[/tex]
P(A) =[tex]\dfrac{1}{500}[/tex] and P(B) = [tex]\dfrac{1}{800}[/tex]
Since A and B are independent events.
According to question,
[tex]P(A\cap B)=P(A).P(B)\\\\P(A\cap B)=\dfrac{1}{500}\times \dfrac{1}{800}\\\\P(A\cap B)=\dfrac{1}{40000}[/tex]
Hence, probability that two people have the same number of repeats in both location A and location B is [tex]\dfrac{1}{40000}[/tex]
