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
