Question

25% of American households have only dogs (one or more dogs) 15% of American households have only cats (one or more cats) 10% of American households have dogs and cats (one or more of each) 50% of American households do not have any dogs or cats. A researcher randomly samples 10 American households. Find the probability that exactly 3 have only dogs, 2 has only cats, 1 has cats and dogs, and 4 has neither cats or dogs. Round your answer to 4 decimal places.

Answer 1

Answer:

a) P=0.2503

b) P=0.2759

c) P=0.3874

d) P=0.2051

Step-by-step explanation:

We have this information:

25% of American households have only dogs (one or more dogs)

15% of American households have only cats (one or more cats)

10% of American households have dogs and cats (one or more of each)

50% of American households do not have any dogs or cats.

The sample is n=10

a) Probability that exactly 3 have only dogs (p=0.25)

$P(x=3)=\binom{10}{3}0.25^30.75^7=120*0.01563*0.13348=0.25028$

b) Probability that exactly 2 has only cats (p=0.15)

$P(x=2)=\binom{10}{2}0.15^20.85^8=45*0.0225*0.27249=0.2759$

c) Probability that exactly 1 has cats and dogs (p=0.1)

$P(x=1)=\binom{10}{1}0.10^10.90^0=10*0.1*0.38742=0.38742$

d) Probability that exactly 4 has neither cats or dogs (p=0.5)

$P(x=4)=\binom{10}{4}0.50^40.50^6=210*0.0625*0.01563=0.20508$