I am having a hard time trying to figure out this question. Any help is appreciated. Thanks!
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?
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 Sn = 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
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
To solve this problem, we can use the concept of probability and the binomial distribution. The binomial distribution models the number of successes (in this case, receiving a PS3) in a fixed number of independent trials (opening gifts), where each trial has the same probability of success (p=0.5).
The question asks for the minimum number of gifts you need to open in order to have a 99% confidence of receiving your first PS3. In other words, we need to find the number of trials that would give us a cumulative probability of at least 99% of getting a PS3.
We can use the cumulative distribution function (CDF) of the binomial distribution to calculate the probability of getting at least one success in a given number of trials. The formula for the CDF is:
C(k, n) = P(X ≤ k) = ∑(i=0 to k) C(i, n) * p^i * (1-p)^(n-i)
Where:
- k is the number of successful outcomes (at least 1 in this case)
- n is the number of independent trials (the number of gifts you open)
- C(i, n) is the binomial coefficient, or the number of combinations of choosing i successes from n trials
- p is the probability of success (getting a PS3)
- (1-p) is the probability of failure (not getting a PS3)
To calculate the number of gifts needed to have a 99% confidence of receiving your first PS3, we need to find the smallest value of n (the number of trials) that satisfies the equation:
C(1, n) = P(X ≤ 1) = ∑(i=0 to 1) C(i, n) * p^i * (1-p)^(n-i) ≥ 0.99
You can calculate this equation for different values of n until you find the smallest value that satisfies the inequality. This can be done manually, but it is more convenient to use a calculator or statistical software that can perform these calculations.
For example, using software like Excel, you can use the BINOM.DIST function to calculate the cumulative probability. By trying different values for n, you can find that opening 7 gifts would give you a cumulative probability of approximately 0.996, which is greater than 0.99. Thus, you can say with 99% confidence that you will receive your first PS3 after opening 7 gifts.
Note that this calculation assumes that the gift outcomes are independent, and the probability of receiving a PS3 is fixed at 0.5 for each gift. This is an idealized scenario and may not perfectly represent real-life situations, but it can be a useful approximation.