Question

A fast-food chain randomly attaches coupons for prizes to the packages used to serve french fries. Most of the coupons say “Play again,” but a few are winners. Seventy-five percent of the coupons pay nothing, with the rest evenly divided between “Win a free order of fries” and “Win a free sundae.” (a) If each member of a family of three orders fries with her or his meal, what is the probability that someone in the family is a winner?
(b) What is the probability that one member of the family gets a free order of fries and another gets the sundae? The third wins nothing.
(c) The fries normally cost $1 and the sundae $2. What are the chances of the family winning $5 or more in prizes?

Answer 1

Solution

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then

probability mass function (pmf) of X is given by

p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n …………………..(1)

Now to work out the solution,

Seventy-five percent of the coupons pay nothing, with the rest evenly divided between “Win a free order of fries” and “Win a free sundae.

=> probability of winning a prize = 1 – 0.75 (i.e., 75%) = 0.25 ……………………………….. (3)

and of this,

probability of winning a free order of fries = 0.25/2 = 0.125 ………………………………….. (4)

and probability of winning a sundae = 0.25/2 = 0.125 ………………………………………….. (5)

let X = Number of members of a family of 3 winning a prize. Then,

X ~ B(3, 0.125) for free order of fries and for sundae separately………………………………….. (6)

X ~ B(3, 0.25) for a prize …………………………………………………………………………. (7)

Part (a)

Probability that someone in the family is a winner

= 1 – P(none is a winner)

= 1 – P(X = 0) [vide (7)]

= 1 - (³C₀)(0.25⁰)(0.75)³ [vide (1)]

= 1 – 0.421875

= 0.5781 Answer

Part (b)

Probability that one member of the family gets a free order of fries, another gets the sundae and the third wins nothing

= 6 x 0.125 x 0.125 x 0.75 [(3), (4) and (5), factor 6 is due to the 3! Arrangements among the three members.]

= 0.072 Answer

Part (c)

Given fries normally cost $1 and the sundae $2, the chances of the family winning $5 or more in prizes

= chances of two members winning a sundae each and the third winning a free fries or all three members winning a sundae each

= {(³C₂)(0.125)²(0.125)} + (0.125)³

= 0.0078125 Answer

DONE