I am having a hard time trying to figure out this question. Any help is appreciated. Thanks!

Question

Suppose that every time you open a Christmas gift, there is a probability, p=0.5 that the gift will be a PS3. Given this probability, how many gifts would you need to open to say with 99% confidence that you will receive your first PS3?

Answer 1

You could get the present on the first try
--->.5
you could get it on the second try ---> .5^2

If S is success and F is failure

we have the following

.5 + .5^2 + .5^3 + ... + .5^n = .99

the left side is a geometric series with
a = .5
r = .5
n = ?
and S_n = 0.99

.99 = .5(1- .5^n)/(1-.5)
.99 = 1 - .5^n
.5^n = .01
n = log(.01)/log(.05)
n = 6.6 , but n is the number of tries, which must be whole number

if n= 6, S(6) = .98, not 99%
if n=7 , S(7) = .992 > 99%

so to be 99% certain of getting his present, he/she should open at most 7 presents

Answer 2

I am going to assume that there is an unlimited supply of gifts and that the probability of getting a PS3 with each opening is 0.5 regardless of previous results. Other situations are possible, such as only two gifts, with one of them a PS3. In that case the answer is obviously 2.

After N openings, the probability of getting no PS3 is (1/2)^N, so the probability of getting one or more PS3's is 1 - (1/2)^N. If N= 7, that probability is 0.992. If N = 6, the probability of getting one or more PS3's is 0.984

I would say that the answer is 7 openings are required. In this case, the probabilities of getting numbers of PS3's from 0 to 7 are:

P(0) = (0.5)^7 = 0.008
P(1) = 7!/(6!*1!)*(0.5^7) = 0.055
P(2) = 7!/(5!*2!)*(0.5^7) = 0.164
P(3) = 7!/(4!*3!)*(0.5^7) = 0.273
P(4) = 7!/(3!*4!)*(0.5^7) = 0.273
P(5) = 0.164
P(6) = 0.055
P(7) = 0.008