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Urn I contains $2$ white and $3$ blue balls. Urn II contains $3$ white and $4$ blue balls. Randomly pick a ball from Urn I and put it into Urn II, and then a ball is picked randomly from Urn II. What is the probability that the second pick is blue

$A=\{Second \ pick \ is \ blue\}$

$B=\{First \ pick \ is \ blue\}$

The way I did it is wrong, however, I cannot pinpoint my faulty logic

$P(A)=P(A \vert B)+P(A \vert B^{\mathsf{c}}) = \frac{P\left(A \cap B\ \right)}{P\left(B\right)}+ \frac{P\left(A \cap B^{\mathsf{c}}\ \right)}{P\left(B^{\mathsf{c}}\right)}= \frac{\frac5 8}{\frac 3 5}+\frac{\frac4 9}{\frac 2 5}=\frac {155}{72}$ which is obviously incorrect.

The correct answer multiplies the numerator and denominator, instead of dividing. Why?

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2 Answers 2

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By law of total probability,

$$P(A)=P(A|B)\color{red}{P(B)}+P(A|B^c)\color{red}{P(B^c)}$$

You miss out two of the terms.

Note that $P(A|B)P(B)=P(A \cap B)$

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  • $\begingroup$ Why does the law of total probability apply here? $\endgroup$
    – Allan
    Commented Oct 18, 2017 at 4:22
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    $\begingroup$ it is used when you want to condition on something, in this case $B$ and $B^c$. $\endgroup$
    – Siong Thye Goh
    Commented Oct 18, 2017 at 4:23
  • $\begingroup$ Follow up: then isn't $P(A \vert B)P(B)=\frac{P\left(A \cap B\ \right)}{P\left(B\right)}P(B)$ when we need $P\left(A \cap B\ \right)P(B)$? $\endgroup$
    – Allan
    Commented Oct 18, 2017 at 4:36
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    $\begingroup$ To solve the problem, you just need $P(A)=P(A|B)P(B)+P(A|B^c)P(B^c)$, I am writing the extra line as to many $P(A) = P(A \cap B) + P(A \cap B^c)$ is obvious while the first equation is not too obvious to many. $\endgroup$
    – Siong Thye Goh
    Commented Oct 18, 2017 at 4:38
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To attack the problem is not to condition, but to us the fact that if $E$ and $F$ are mutually exclusive events, then $P(E\cup F) = P(E) + P(F)$. Using this approach, note that $A = (A\cap B) \cup (A\cap B^{\complement})$. It this follows that \begin{align} P(A) &= P(A\cap B) + P(A\cap B^{\complement}) \\ &= P(A \mid B) P(B) + P(A \mid B^{\complement}) P(B^{\complement}). \end{align} Now, note that if the first ball selected is blue, then the probability of the second ball being blue is $\frac{5}{8}$, since there will be 8 balls in Urn II, 5 of which will be blue. Note also that the probability of the first ball being blue is $\frac{3}{5}$. Thus $$ P(A \mid B) P(B)= \frac{5}{8} \cdot \frac{3}{5} = \frac{3}{8}. $$ By similar reasoning, $$ P(A \mid B^{\complement}) P(B^{\complement}) = \frac{4}{8} \cdot \frac{2}{5} = \frac{1}{5}. $$ Combining these, we obtain $$ P(A) = P(A \mid B) P(B) + P(A \mid B^{\complement}) P(B^{\complement}) = \frac{3}{8} + \frac{1}{5} = \frac{15+8}{40} = \frac{23}{40}. $$

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    $\begingroup$ @Allan $\{A\cap B\}$ is the event of picking one from the 3 blue balls among 5, transfer it to the other urn then pick one from the now 5 blue balls among 8. $\mathsf P(A\cap B)=\tfrac 35\tfrac 58 = \tfrac 38$ $\endgroup$
    – Graham Kemp
    Commented Oct 18, 2017 at 4:55

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