please help me solve it by steps

Question

One of two urns is chosen at random with one as likely to be chosen as the other. Then a ball is withdrawn from the chosen urn. Urn 1 contains 1 white and 5 red​ balls, and urn 2 has 4 white and 2 red balls. If a red ball is​ drawn, what is the probability that it came from urn 2​? ​(Type an integer or a fraction. Simplify your​ answer.)

Answer 1

Urn 1 has 1 white and 5 red balls

Urn 2 has 4 white and 2 red balls

if a red ball is drawn, what's the probability it came from Urn 2?

use Bayes theorem

P(R/U2)P(U1= P(U2/R)(P(R)

P(R/U2) = 2/6=1/3

P(U2) = 1/2

P(R)=7/12

(1/3)(1/2) = P(U2/R)(7/12)

P(U2/R) = (1/6)/(7/12) = 7/72 = about 10% chance that red ball is from Urn 2

Answer 2

The probability of drawing a red ball from Urn 1 is 5/6, since there is 1 white ball and 5 red balls in Urn 1.

The probability of drawing a red ball from Urn 2 is 2/6, since there are 4 white balls and 2 red balls in Urn 2. (2/6) = (1/3), BUT it will be better to leave it as 2/6 so that we already have a common denominator.

Since no white balls were drawn, we can consider only the probability that the red ball came from Urn 2 compared to the probability that the red ball came from Urn 1.

It was given that it was equally likely that the drawn ball came from either Urn.

So the ratio of the probabilities give us the probability that the red ball came from Urn 2. That is:

(2/6) / [ (5/6) + (2/6) ] = (2/6) / [(7/6)] = 2/7

NOTE that once we know the drawn ball was red, we know that 2/6 = 1/3 of the time on average it came from Urn 2, and 5/6 of the time it came from Urn 1. So we expect that 2 times out of 7, the red ball will come from Urn 2.

Do you understand this logic, or is there a step that needs more explanation? If so, which step?

Answer 3

Joanna,

You're definitely on a Bayes' Theorem problem. Let's review what that is starting with definition of conditional probability (probability of A given B):

P(A | B) = P(A ∩ B) / P (B)

Multiply the P(B) to both sides and we get:

P(A | B) P(B) = P(A ∩ B) or (flipping around)

P(A ∩ B) = P(A | B) P(B)

Should read intuitively, the probability of A and B happening is same as probability of A happening given be times the probability of B happening. Example: probability of flipping two heads in a row on a coin (1/2*1/2) is same as flipping a heads (1/2) given the previous was a heads (1/2) so again (1/2*1/2).

Now cleverly let's plug this into P(B|A) (remember P(A ∩ B) = P(B ∩ A)):

P(B | A) = P(B ∩ A) / P(A)

P(B | A) = P(A | B) P (B) / P (A)

So let's collect some stuff:

P( white ball | urn 1) = 1/6

P( red ball | urn 1) = 5/6

P( white ball | urn 2) = 4/6

P( red ball | urn 2) = 2/6

Key here in wording from "One of two urns is chosen at random with one as likely to be chosen as the other." is:

P( picking from urn 1) = 1/2

P(picking from urn 2) = 1/2

Now finally what is the question: " If a red ball is​ drawn, what is the probability that it came from urn 2​?" Let's write that as a conditional:

P(urn 2 | red ball)

Using Bayes' Theorem ( P(B | A) = P(A | B) P (B) / P (A) ) and plugging in everything we have we get:

P(urn 2 | red ball) = P(red ball | urn 2) P(urn 2) / P(red ball)

P(urn 2 | red ball) = (2/6) * (1/2) / P(red ball)

But what's P(red ball)? Well let's use total probability P(red ball) = P(red | urn 1) P(urn 1) + P(red | urn 2) P(urn2). This just says the probability of an event can be rewritten as sum of all the individual conditions. SO for us P(red) = 5/6 * 1/2 + 2/6 * 1/2 = 7/12

So our final answer is:

P(urn 2 | red ball) = (2/6) * (1/2) / (7/12)

P(urn 2 | red ball) = (2/12) / (7/12)

P(urn 2 | red ball) = 2/7

This checks out intuitively since urn 2 has a lot less red balls than urn 1. A way to think of this is just if I have just a ball of any kind the best guess is 50/50 for urn 1 and urn 2 right? But now that I tell you it's red, you can "update" your probability and it's more likely urn 1. This is a very common way to think of Bayes' Theorem in probability: prior, update, and posterior (i.e. probability, evidence, new probability based on evidence).