Question

4. Given a population composed of dogs (40%) and cats (60%), in a sample of size of 10

Find the probability of 3 cats.

Find the probability of fewer than 3 dogs.

Find the probability of between 4 and 7 cats.

Find the probability of 2 cats and 2 dogs.

Answer 1

This can be used as a Binomial distribution.

We know that:

The probability of choosing a dog, that is, the probability of success, p is 0.40.

The probability of choosing a cat, that is, the probability of failure, q is 0.60.

The sample size,n = 10

We know that for a Binomial distribution with X~B(n,p),

P(X=r)= nCr * p^r * q^n-r

The probability of 3 cats, or 7 dogs is the case when r=7.

Substituting r=7 in the above equation, we get

P(X=7)= 0.042467328

The probability of fewer than 3 dogs will be when r=0,r=1 and r=2.

Substituting r=0,1,2 in the above equation, we get

P(X=0)= 0.0060466176

P(X=1)= 0.040310784

P(X=2)= 0.120932352

P(X<3)= P(X=0)+P(X=1)+P(X=2)

= 0.1672897536

P(r=6,5,4,3)

P(X=6)= 0.111476736

P(X=5)= 0.2006581248

P(X=4)= 0.250822656

P(X=3)= 0.214990848

P(2<X<7)= P(X=6)+P(X=5)+P(X=4)+{(X=3)

= 0.7779484

Since we cannot have 2 cats and 2 dogs in a sample size of 10 (as the sum is not 10), this result is not possible.

Hence the probability is 0.