Probability?

Question

1)You have three boxes. One contains two pennies, one contains a penny and a dime and the third contains two dimes. You randomly select a box and randomly pull out a penny. Is the following statement correct? There is a 50% chance that the other coin in that box is a penny. Explain.

The correct answer is no. There is something wrong with the statement, but I don't know what.


2) You are told that there is a 25% chance of it raining on any given day during August. You are also told there is a 50% chance of rain if it rained the day before and a 16.66% of rain if it did not rain the day before. Determine if these probabilities are all consistent.

I don't know. I am thinking that since there is a 25% chance of raining on any given day, it will rain for about 25% of the month. The chance of rain should be greater then if it didn't rain the day before so I don't think the probabilities here are sensible. Am I correct?

Answer 1

I will use the notation P(A | B) to mean the probability of event A occuring given the probability of event B occuring. One way to calculate this (besides intuitive reasoning is)

P(A | B) = P(both A and B) / P(B),

where the forward slash naturally denotes division.

=====PROBLEM 1=====

Let's label the boxes B1, B2, and B3.

B1 contains two pennies, p1 and p2.

B2 contains two dimes, d1 and d2.

B3 contains the penny p3 and the dime d3.

The probability of choosing any given coin is 1/6 (it's equally likely regardless of the coin), even though we can't tell p1, p2, and p3 apart, and we can't tell d1, d2, and d3 apart, if we just happen to draw one coin. This is because each box is chosen with 1/3 probability, then a given coin from the box with 1/2 probability, and 1/3 * 1/2 = 1/6.

Since we know P(p1) = P(p2) = P(p3), we know that if we draw a coin and it happens to be a penny, we're equally likely to draw any of the three (once again, though, we won't necessarily know which we draw). More specifically, we have

P(p1 drawn | penny drawn) = 1/3,

P(p2 drawn | penny drawn) = 1/3,

P(p3 drawn | penny drawn) = 1/3.

Notice that if p1 or p2 is drawn there will be another penny in the box, but if p3 is drawn there won't be.

So finally

P(coin drawn from B1 | penny drawn) = 1/3 + 1/3 = 2/3 ≠ 1/2.

=====PROBLEM 2=====

It's possible I'm mistaken, but I'm fairly sure it's consistent. I'm also making an assumption that by 16.66% they actually mean 1/6 (as 16.66% is very close to 1/6, and I think they're just writing it this way to drop off the infinite string of 6's, BUT, it's odd that they didn't round it to 16.67%).

Anyway, here goes.

Let's assume the probability that it rains on August 1st is independent on whether or not it rained on July 31st, as I think the data only applies to August. Even if this isn't true, I don't think it matters, but it makes approaching the problem simpler.

I'll use the notation R1 to mean it rained on the first, R2 to mean it rained on the second, and so on. I'll use N instead of R to mean that it didn't rain.

P(R1) = 1/4 [given]

P(N1) = 1 - 1/4 = 3/4

P(R2) = P(R2|R1) * P(R1) + P(R2|N1) * P(N1)
<=>
P(R2) = 1/2 * 1/4 + 1/6 * 3/4 = 1/4

Hence, so far, everything seems consistent. August 2nd is a given day of August, and it should rain with 1/4 probability on that day according to the first rule. The second rule doesn't contradict the first in the case of August 2nd. Now in the "math" above just replace R1 and N1 with R2 and N2 and and R2 and N2 with R1 and N3 (respectively), and you'll be proving that everything works for the third too and so on. But let's use a trick called "induction" to prove it (probably not necessary, but it's good practice).

Let n be some arbitrary day between August 1st and 30th and let m be the day after (notice that m will always be in August as well, because even if n is the 30th, m is the 31st).

Assume P(Rn) = 1/4.

P(Rn) = 1/4 [given]

P(Nn) = 1 - 1/4 = 3/4

P(Rm) = P(Rm|Rn) * P(Rn) + P(Rm|Nn) * P(Nn)
<=>
P(Rm) = 1/2 * 1/4 + 1/6 * 3/4 = 1/4

Hence as long as it rains with 1/4 probability on Day n it will rain with a 1/4 chance on Day m using either rule.

Since we know on Day 1 we have a 1/4 chance of rain it follows by "induction" (and the fact we just proved above) that since it works for the 1st it works for the 2nd, since it works for the 2nd it works for the 3rd, since it works for the 3rd it works for the 4th, and so on...

Hope that helps!

Edit your question to ask for clarifications if you're lost on anything.

Answer 2

1) you have 3 boxes that contain:
[p1 p2] [p3 d1] [d2 d3]
you've picked a box and pulled out a penny (but don't know which).
if p1, other coin is p2
if p2, other coin is p1
if p3, other coin is d1
of the 3 possibilities, 2 have the other coin a penny, so probability is 2/3 or 66.7%.

2) I'll have to think about this one, and it's too late tonight to do it. Good luck.