Applying Bayes Theorem in a word problem involving drawing balls in succession

Question

I just got introduced to Bayes theorem, and tried solving the following question:

Two balls are drawn in succession from a jar containing 3 red balls and 4 white balls. What is the probability that the first ball was white given the second ball was red.

I used the formula given in the book implicitly using the following tree diagram,

And, then I got: $$\frac{\frac{1}{2} \cdot \frac{2}{6}}{\frac{1}{2} \cdot \frac{2}{6} + \frac{1}{2} \cdot \frac{3}{6}}$$

where $\frac{1}{2}$ represents the probability of either picking a white or red ball when we draw the first ball, $\frac{2}{6}$ represents the probability of drawing a red ball when we draw the second ball, and $\frac{3}{6}$ represents the probability of drawing a white ball when we draw the second ball.

Did I do it correctly?

Answer 1

No since there are no longer 7 balls after you take the first ball out.

$$\begin{split}P(W^1|R^2)&=\frac{P(W^1R^2)}{P(W^1R^2)+P(R^1R^2)}\\ &=\frac{\color{green}{\frac 47}\cdot \frac 3 {\color{red}6}}{\color{green}{\frac 47}\cdot \frac 3 {\color{red}6}+\color{blue}{\frac 37\cdot\frac 26}}\end{split}$$

The blue part is if you draw red on the first draw.

Answer 2

There are $7$ balls, $4$ white and $3$ red. So, it should be,

$ \displaystyle \frac{P(WR)}{P(WR) + P(RR)} = \frac{\frac{4}{7}\cdot\frac{3}{6}}{\frac{4}{7} \cdot \frac{3}{6} + \frac{3}{7} \cdot \frac{2}{6}}$

The denominator is probability of second ball being red - which can be either both balls red or first ball White and second ball red.

The numerator is probability of first ball being white and second being red.

Answer 3

No. You need : $$\dfrac{\mathsf P(R_2\mid W_1)\mathsf P(W_1)}{\mathsf P(R_2\mid W_1)\mathsf P(W_1)+\mathsf P(R_2\mid R_1)\mathsf P(R_1)}$$

So you have the right structure but the wrong numbers.

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If the second ball is red the first ball will have been one among the other balls: two of which were red, four of which were white.

The probability that the first ball is white given that the second ball is red is equal to the probability that the second ball is white given that the first is red.