1. ## 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.

2. Hello, flywithme!

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

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