# Conditional Probability question

• Jan 21st 2010, 11:06 PM
flywithme
Conditional Probability question
Bag M has 5 white balls and 2 red balls. Bag N contains 3 white balls and 4 red balls.
If a white ball is selected from Bag N, What is the probability that a red ball was transferred from Bag M to Bag N?
Please give the answer in any form you like, fractions, decimals, etc. Also, step by step instruction is also very much appreciated.
• Jan 22nd 2010, 07:47 AM
Soroban
Hello, flywithme!

You left out part of the problem.
I'll take a guess at what's missing.

Quote:

Bag M has 5 white balls and 2 red balls.
Bag N contains 3 white balls and 4 red balls.

A ball is randomly chosen from Bag M and moved to Bag N.

If a white ball is selected from Bag N, what is the probability
that a red ball was transferred from Bag M to Bag N?

This is Conditional Probability, requiring Bayes' Theorem.

. . $P(\text{R moved }|\text{ W from N}) \;=\;\frac{P(\text{R moved }\wedge\text{ W from N})} {P(\text{W from N})}$ .[1]

We consider the two cases . . .

[1] White moved from Bag M: . $P(\text{W moved}) \,=\,\frac{5}{7}$
. . Then bag N has 4W, 4R: . $P(\text{W from N}) \,=\,\frac{4}{8}$
. . Then: . $P(\text{W moved }\wedge\text{ W from N}) \;=\;\frac{2}{7}\cdot\frac{4}{8} \;=\;\frac{10}{28}$

[2] Red moved from Bad M: . $P(\text{R moved}) \,=\,\frac{2}{7}$
. . Then Bag N has 3W, 5R: . $P(\text{W from N}) \,=\,\frac{3}{8}$
. . Then: . $P(\text{R moved }\wedge\text{ W from N}) \;=\;\frac{2}{7}\cdot\frac{3}{8} \;=\;\frac{3}{28}$ .[2]

Hence: . $P(\text{W from N}) \:=\:\frac{10}{28} + \frac{3}{28} \:=\:\frac{13}{28}$ .[3]

Substitute [2] and [3] into [1]: . $P(\text{R moved }|\text{W from N}) \;=\;\frac{\dfrac{3}{28}}{\dfrac{13}{28}} \;=\;\frac{3}{13}$