Question

I haven’t learned random variable so it needs to be more basic than that. Just using P(A) and so on

60% of animals are cats (10% brown) and 40% are dogs (1 % brown) If 10 are randomly chosen is the probability that at least 2 will be dogs. please use the P(A) method.

Answer 1

It is given that 60% of animals are cats and the remaining 40% are dogs.

So, the probability that a randomly chosen animal is a dog is 0.4

Next, we randomly choose 10 animals.

Probability that no dog gets selected = Probability that all are cats = 0.6^10 = 0.00604

Probability that only 1 dog is selected = ¹⁰C₁*(0.4^1)*(0.6^9) = 10*(0.4^1)*(0.6^9) = 0.0403

So,

Probability that less than 2 dogs are selected = Probability that no dog is selected + Probability that only 1 dog is

selected = 0.00604+0.0403 = 0.04634

Using the complement rule:

P(A) = 1 - P(A^c)

So,

Probability that atleast 2 dogs are selected = 1 - Probability that less than 2 dogs are selected = 1-0.04634

= 0.95366