Thread: Conditional Probability Brain Teaser

1. Conditional Probability Brain Teaser

A sock contains 2 red marbles and an unknown number of blue marbles. Tom places a new marble in the sock. Jason puts his hand in the sock and pulls out a red marble.

What is the probability that the marble Tom put in the sock was red?

I've been trying to approach this with Bayes' Theorem but not making any progress:

P(A l B) = P(B l A) * P(A) / P(B)

I'm confused as to what A and B would represent in the problem.

Appreciate the help, thanks!

2. Originally Posted by Penguins
A sock contains 2 red marbles and an unknown number of blue marbles. Tom places a new marble in the sock. Jason puts his hand in the sock and pulls out a red marble.

What is the probability that the marble Tom put in the sock was red?

I've been trying to approach this with Bayes' Theorem but not making any progress:

P(A l B) = P(B l A) * P(A) / P(B)

I'm confused as to what A and B would represent in the problem.

Appreciate the help, thanks!
Let the number of blue marbles in the sock be x.

Draw a tree diagram. The first two branches are the events Tom putting in a blue marble or a red marble. The next branches are the events of pulling out a red marble or a blue marble.

I get $\frac{\frac{3}{2(3 + x)}}{\frac{3}{2(3 + x)} + \frac{1}{3 + x}} = \frac{3}{5}$.

3. I don't understand how you set up the problem like you did. What do the terms in your numerator and denominator represent?

Here is what I get:

P(A l B) = (P(B l A) * P(A)) / P(B)

A is Tom putting a red marble in the sock.
B is Jason taking a red marble out of the sock.

P (B l A) = 3 / (x+3)
P (A) = unknown, let it be p
P (B) = p*(3/(x+3)) + (1-p)*(2/(x+2))

So

P(A l B) = (P(B l A) * P(A)) / P(B) = (3p/(x+3)) / (p*(3/(x+3)) + (1-p)*(2/(x+2)))

I end up with 3/2 p. I know I did something wrong, but I'm not sure what.

4. Hello, Penguins!

I agree with Mr. F

A sock contains 2 red marbles and an unknown number of blue marbles.
Tom places a new marble in the sock.
Jason puts his hand in the sock and pulls out a red marble.

What is the probability that the marble Tom put in the sock was red?
We want: . $P(\text{add Red }|\text{ draw Red}) \;=\;\frac{P(\text{add Red }\wedge\text{ draw Red})}{P(\text{draw Red})}$ .[1]

The sock contains: 2 Red and $b$ Blue marbles.

There are two cases to consider:

[1] A Red marble is added.
. . .The sock contains: 3 Red and $b$ Blue.
. . . . Then: . $P(\text{add Red }\wedge\text{ draw Red}) \:=\:\frac{3}{b+3}$ .[2]

[2] A Blue marble is added.
. . .The sock contains: 2 Red and $b+1$ Blue.
. . . . Then: . $P(\text{add Blue }\wedge\text{ draw Red}) \:=\:\frac{2}{b+3}$

Hence: . $P(\text{draw Red}) \;=\;\frac{3}{b+3} + \frac{2}{b+3} \;=\;\frac{5}{b+3}$ .[3]

Substitute [2] and [3] into [1]: . $P(\text{add Red }|\text{ draw Red}) \;\;=\;\; \frac{\dfrac{3}{b+3}} {\dfrac{5}{b+3}} \;\;=\;\;\frac{3}{5}$

5. Originally Posted by Soroban
Hello, Penguins!

I agree with Mr. F

We want: . $P(\text{add Red }|\text{ draw Red}) \;=\;\frac{P(\text{add Red }\wedge\text{ draw Red})}{P(\text{draw Red})}$ .[1]

The sock contains: 2 Red and $b$ Blue marbles.

There are two cases to consider:

[1] A Red marble is added.
. . .The sock contains: 3 Red and $b$ Blue.
. . . . Then: . $P(\text{add Red }\wedge\text{ draw Red}) \:=\:\frac{3}{b+3}$ .[2]

[2] A Blue marble is added.
. . .The sock contains: 2 Red and $b+1$ Blue.
. . . . Then: . $P(\text{add Blue }\wedge\text{ draw Red}) \:=\:\frac{2}{b+3}$

Hence: . $P(\text{draw Red}) \;=\;\frac{3}{b+3} + \frac{2}{b+3} \;=\;\frac{5}{b+3}$ .[3]

Substitute [2] and [3] into [1]: . $P(\text{add Red }|\text{ draw Red}) \;\;=\;\; \frac{\dfrac{3}{b+3}} {\dfrac{5}{b+3}} \;\;=\;\;\frac{3}{5}$

I see what you're doing, but isn't it a problem that you're assuming the probability of Tom putting a red marble in the sock to be 1/2? I set up the problem using p as this probability and unfortunately made a careless mistake in my previous post, but get the answer to the question as (3p)/(2+p). When p = 1/2, the answer is 3/5.

I'm curious as to whether there is a rule in a situation like this as to whether lacking more information it is assumed that all possibilities have equal likelihood?

Thanks for the replies!

6. Originally Posted by Penguins
I see what you're doing, but isn't it a problem that you're assuming the probability of Tom putting a red marble in the sock to be 1/2? I set up the problem using p as this probability and unfortunately made a careless mistake in my previous post, but get the answer to the question as (3p)/(2+p). When p = 1/2, the answer is 3/5.

I'm curious as to whether there is a rule in a situation like this as to whether lacking more information it is assumed that all possibilities have equal likelihood?

Thanks for the replies!
If you're expected to get a numerical answer then you have to make an assumption regarding a priori knowledge. The best one (for various reasons which I won't get into here) is a uniform prior distribution.