Question

You collect baseball cards and buy sealed packs from the grocery store and clearly have no idea which cards are inside (nor if they will be valuable). Suppose 1 in every 250 cards is valuable and a package contains 10 cards which are selected independently and randomly at the factory.
(a) If you purchase a package, what is the probability that there are no valuable cards? t you pucha a packae, wst s the probbiiythat thee is a east I valuable curd
(c) What is the expected number of valuable cards in the package?
(d) Suppose you buy 5 packages, what is the probability that none of these packages have valuable cards in them?
(e) Suppose you keep buying packages until you get 1 valuable card. What is the probability that you buy 2 packages?
(f) Suppose you desire to have 3 valuable cards in your collection. What is the probability that it takes 20 packages to accomplish this? On average, how many packages would it take to accomplish this?

Answer 1

Answer:

(a) 0.9607

(c) 0.04

(d) 0.8184

(e) 0.074

(f) 0.0853/e^0.8;

750 cards

Step-by-step explanation:

Let the probability of success of a valuable card be p

p= 1/250 = 0.004,

q = 1-p = (249/250)

(a) There are 10 cards in one pack

The probability that there are no valuable card is given by:

Pr(X= 0) = 10C0 × (1/250)^0 ×(249/250)^10

10C0 = 10 combination 0= 10!/10!0!

=1

Pr (X= 0) = (249/250)^10

Pr (X=0) = 0.9607

(c) Expected number of valuable card is given by:

E(X) = np , n= 10, p= 1/250

E(X) = 10 × 1/250 = 1/25 = 0.04

(d) 5 packages means 5 × 10= 50 cards

Pr(X=0) = 50C0 × (1/250)^0 ×(249/250)^50

50C0 = 50!/50! ×0!=1

Pr (X=0) = (249/250)^50

Pr (X= 0) =(0.996)^50

Pr (X=0) = 0.8184

(e) For this case we use the Binomial Distribution

2 packages 2 × 10 = 20 cards

n = 20 , x = 1

We use:

P(X=1) = 20C1 × (1/250)^ 1 × (249/250)^19

Pr (X=1) = 20C1 × (1/250)^1 × (249/250)^19

Pr (X=1) = 20 × (1/250)¹ × (249/250)^19

Pr (X= 1 )= 0.074

(f) 20 packages = 200 cards

For this we apply Poisson distribution

P(X=3) = (e ^-h × h ^x) / x!

Where h = np = 200/250 = 0.8

P(X=3) = e^-0.8 × (0.8)^3 / 3!

P(X=3) = 0.991 × 0.512/6

P(X=3) = 0.0846

3 = n p

n = 3/p = 3 /0.04

n = 750 cards