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?

(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

nope.   happens to me as well.

-aparri