"A batch of 350 raffle tickets contains three winning tickets. You buy five tickets. What is the probability that you have (a) no winning tickets, (b) all of the winning tickets, (c) at least one winning ticket, and (d) at least one nonwinning ticket?"
I would like the answers to the questions above, but I'd more like to know which sample is n and which is p.
The probability of success, p, is 3/350 = .00857 so (1-p) = .99142
Your number of trials is five, n=5
The probability of k successes in n trials is [binomial coeff n,k) p^k (1-p)^(n-k)
row five of Pascal's triangle (my easy way to get binomial coefficients rather than doing n!/[k!(n-k)!] ) is
1 5 10 10 5 1
so for example the probability of ZERO successes in your five trials is:
(1) (.00857)^0 (.99142)^5
=(1)(1)(.958)
=.958
Try the rest after checking my arithmetic carefully.
Ok. That helped me out a lot. I have another one, if your willing to give direction.
One in five people in the U.S. own individual stocks. In a random sample of 12 people, what is the probability that the number owning individual stocks is (a) exactly four, (b) at least four, and (c) more than four.
So, in this case, my p is .20, correct? and my n is 12?
However, I am not sure the binomial distribution applies here. Your probability is not constant as you draw winning tickets. The trials are not independent because if you get one winning ticket, your chances of getting a second go down. There are only two left in the pot if you draw one winner.
Therefore question a, NO winning tickets. Initially you have 350-3 =347losers, each time you take a loser out, you have one less:
(347/350)(346/349)(345/348)(344/347)(343/346)
= .958
That works out the same because we did not really change the nature of the population by drawing losers, However if we pull a winner out, things will change
Let's try for all(3) winners in our five trials.
Your chances for the first one are
5(3/350)
Your chances for a second one are
4(2/349)
Your chances for a third and final one are
3(1/348)
Multiply those together to find the chance of getting all three:
=8.47*10^-6 or about 8 out of a million
Now try the rest.
Yes, p = 0.2 and n = 12 for the stock one. In this case the trials are indeed independent so you can use binomial distribution safely.
I am wondering if the chapter you are working on is about binomial distributions if the first problem is really intended to have non-independent trials. Strange
Well the question in the first problem is earlier in the book than in the second. We did learn about the process that you used early on though.
To answer these questions, we need to identify the sample space (n) and the probability (p).
In this case, the sample space (n) refers to the total number of possible outcomes, which is the number of ways you can choose 5 tickets from the batch of 350 raffle tickets. This can be calculated using the combination formula, "nCr" or "n choose r," which is given by:
nCr = n! / (r! * (n-r)!),
where n! denotes the factorial of n.
Applying this formula, we find that the sample space (n) in this case is:
n = 350C5 = 350! / (5! * (350-5)!)
The probability (p) refers to the likelihood of a specific event occurring. To determine the probability, we need to consider the desired outcomes and divide it by the total number of outcomes (sample space). Let's analyze each question and determine the corresponding sample space and probability.
a) "No winning tickets":
The sample space (n) for this question remains the same: n = 350C5.
The probability (p) of not having any winning tickets can be calculated by dividing the number of ways to choose 5 non-winning tickets from the remaining 347 non-winning tickets by the total sample space (n). Mathematically, it can be expressed as:
p = (347C5) / n
b) "All of the winning tickets":
The sample space (n) remains the same: n = 350C5.
The probability (p) of having all the winning tickets is calculated by dividing the number of ways to choose all 3 winning tickets from the remaining 347 non-winning tickets by the total sample space (n). Mathematically, it can be expressed as:
p = (3C3 * 347C2) / n
c) "At least one winning ticket":
The sample space (n) remains the same: n = 350C5.
The probability (p) of having at least one winning ticket can be calculated by subtracting the probability of not having any winning tickets (calculations provided in Part a) above) from 1. Mathematically, it can be expressed as:
p = 1 - (347C5) / n
d) "At least one non-winning ticket":
The sample space (n) for this question also remains the same: n = 350C5.
The probability (p) of having at least one non-winning ticket can be calculated by subtracting the probability of having all the tickets as winning tickets (calculations provided in Part b) above) from 1. Mathematically, it can be expressed as:
p = 1 - (3C3 * 347C2) / n
Please note that you will need to evaluate the combinations (3C3, 347C5, etc.) using factorial calculations.