Practice Questions – Binomial Distribution

Question 1:

Let’s assume I have a big bowl with 12 green marbles, 30 yellow marbles and 15 blue marbles.

• If I pull with replacement 8 marbles, what’s the chance 6 of them are non-green?
• If I pick 10 marbles with replacement, what’s the expected value of the blue marbles I’ve pulled.
• Let’s say I’ve picked 7 marbles with replacement. Given than non-of the marbles was green, what’s the chance 2 were blue?
• Let’s say I’ve picked 5 marbles without replacement. What’s the chance I’ll get at least 2 colors of marbles?

Answer 1:

• This can be modeled as a binomial distribution with n=8 and p=45/57. So that is the answer:

• Since picking the marbles is with replacement, in this case too, it’s also a binomial distribution, with n=10 and p=15/57.

So the expected value is E(x)=np:

• Since in this case it’s with replacement, that too can also be modeled as a binomial distribution with n=7 but the p is different for the 2 parts of the calculation, and so, that’s the calculation in this case:

It can be seen, that in this case, for the numerator, modeling the scenario as a multinomial distribution, was more accurate since it involved more than 3 results (green marbles, blue marbles and yellow marbles).

• In this case, since the choice is without replacement, it can’t be modeled as a binomial distribution. However, to pull at least 2 colors, that means that the probability we’re looking for is 1 minus the probability to pull exactly 1 color. This is the calculation for that:

Question 2:

Let’s assume I’m spending the day outside shooting some hoops.

• If the chance for success is 0.73, and I throw 12 times, what’s the chance I have 6 successful trials?
• If the chance for success is 0.18, and I throw 7 times, what’s the chance I’ll have at least 3 successful trials?
• If the chance for success is 0.75, and I throw 8 times, what are the odds I’ll succeed exactly 4 times but none of it at the 4^th of at the 6^th try?
• If the chance for success is 0.66, and I throw 155 times, what’s the expected values of non-successful throws?

Answer 2:

In all cases discussed here, the relevant probability is binomial with various number of trials and various probabilities for success.

• That’s the probability in this case:

• That’s the probability in this case:

• Let’s assume that x1 is for the 4^th and the 6^th time, and x2 is for all the rest; So, that’s the probability in this case:

• If the chance for success is 0.66, so the chance for failure is 0.34, and so, the expected value would be:

So the expectation is for 51 failures in this set of trials.