HELPPPPPPP

[Centurion1]

Never mind i figured it out. I reached the 1.4% using both methods, i.e. the hypergeometric and the multiplying out manually.

I appreciate the help. Unfortunately I have more to go and other problems i have found issue with.....

Does anyone have any idea on this one?


2. Photocards-R’-Us is a photo card printer. For this year, they have found that 10% of the cards they make are of inferior quality (blurred picture, faded colors, etc.) To attract customers to have their holiday cards printed with them, they are offering a 50% discount to customers who order a box of 50 cards if 10 of the cards in the box are of inferior quality. Manny is going to order a box of 50 Christmas photo cards at Photocards-R’-Us. What is the probability that he will receive a 50% discount on his holiday cards? [10 pts.]


Personally, I thought that it would be like Bayes' but the wording confuses me as well as the application required.

[Montmorency]

Grah!

Ha.

Complicated math speak here. But used simple combinations to get answer way up there.

2. Rephrase: Probability of getting at least 10 defective cards.

Binomial expansion: Summation for 10-50 cards defective. But simpler summation 0-9 cards defective. Call defective a failure.

p = probability success = .9
q = probability failure = .1
n = # times = 50
x = # successes
n - x = # failures


..............(x=50)
(p + q)^50 = Σ _nC_xp^xq^(n-x)
................(x=41)


And subtract from 1 to get probability you want. Can't think of simpler way.

[Centurion1]

hrmmmm wat?

[Centurion1]

wait is this poisson because ive never seen it in that context

[Montmorency]

You've been poisoned?

[Centurion1]

poisson is a formula.

if you could elaborate i would really appreciate it in all seriousness.

[Montmorency]

What is the probability of getting a 50% discount, when a discount is given for 10/50 cards (at least) defective. 10% probability per card of defectiveness.

Sum probabilities of getting 1, 2, 3, 4, 5, 6, 7, 8, or 9 defective cards and subtract from 1. Or sum probabilities of 10, 11, 12...or 50 defective cards.

Use binomial expansion?

General binomial expansion: _nC₀p⁰qⁿ + _nC₁p¹q^n-1 ... + _nC_n-1p^n-1q¹ + _nC_npⁿq⁰