Question

Problem #3: 37% of all pets sold at a pet store are dogs. One day, 9 people arrive at the store and purchase pets. (Note: One per is purchased per person.) (a) What is the probability that exactly 2 of them are dogs? (b) If the store has only 6 dogs, what is the probability that they have sufficient dogs for sale that day? Problem #3(a): Round your answer to 4 decimals Problem #3(b): Round your answer to 4 decimals. Just Save Submit Problem #3 for Grading Problem #3 Attempt #1 Your Answer: 3(a) 3(b) Your Mark: 3(a) 3(b) Attempt #2 3(a) 3(b) 3(a) 3(6) Attempt #3 3(a) 3(b) 3(a) 3(b)

Answer 1

Probability of selling a dog=p=0.37

Probability of selling other than a dog=q=1-0.37=0.63

number of trials=n=9

a)

Number of dogs sold=r=2

Its a case of binomial probability distribution.

We need to find P(r=2)

We know

P(r)=ⁿC_rp^rq^n-r

P(2)=⁹C₂0.37²0.63^9-2

We know

⁹C₂=9!/(2!*(9-2)!)=9*8/2=36

P(2)=36*0.37²*0.63^9-2=0.1941

b)

It means we need to calculate probability that 6 or lower number of dogs are sold.

Plr<6) = 1 - Plr = 7) - Plr = 8) – Pr = 9)

P(r)=ⁿC_rp^rq^n-r

P(7)=⁹C₇0.37⁷0.63^9-7=36*0.37⁷0.63²=0.013564

P(8)=⁹C₈0.37⁸0.63^9-8=9*0.37⁸0.63¹=0.001992

P(9)=⁹C₉0.37⁹0.63^9-9=1*0.37⁹0.63⁰=0.000130

P(r < 6) = 1-0.013564 -0.001992 -0.000130 = 0.984314

Plr<6) = 0.9843