Question

Consider two boxes, one containing $1$black and $1$white marble, the other $2$black and $1$white marble. A $100$Chapter $3$Conditional Probability and Independence box is selected at random, and a marble is drawn from it at random. What is the probability that the marble is black? What is the probability that the first box was the one selected given that the marble is white ?

Short answer

Given that the marble is white, the probability that the first box was chosen is$\approx 33.9\%$.

Step-by-step solution

Given Information

Experiment: Deck of $52$cards is dealt to $4$players ( $13$cards each), call them North, East, South and West.

Find P(B|A)

$A$- North and South together have $8$spades.

$B$- $Easthas3spades$

$P(B|A)=?$

Outcome space $S$contains is every division of those $52$cards into $4$(ordered) groups of $13$

If all events in$S$are considered equally likely, probability of event $A\subseteq S$is:

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

where $\left|X\right|$denotes the number of elements in $X$.

The number of elements in $S$here is $\left|S\right|=\left(\begin{array}{c}52\\ 13\end{array}\right)\left(\begin{array}{c}39\\ 13\end{array}\right)\left(\begin{array}{l}26\\ 13\end{array}\right)=\frac{52!}{(13!{)}^4}$Choosing cards for the North, East, and South, with the remaining cards going to the West

The formula for conditional probability (for $A,B$events) is:

$P(B\mid A)=\frac{P\left(AB\right)}{P\left(A\right)}$

Step 3:.Determine the number of elements in A

Choose the cards for North and South first, then the eight spades. $\left(\begin{array}{c}13\\ 8\end{array}\right)$ways, and the remaining $18$cards from the non spades in $\left(\begin{array}{l}39\\ 18\end{array}\right)$ways. Then determine the Norths cards from those $26$in $\left(\begin{array}{l}26\\ 13\end{array}\right)$ways.

Whichever choices are done in choosing the cards for the North and South, the remaining 26 cards can be distributed among East and West in any of the $\left(\begin{array}{l}26\\ 13\end{array}\right)$ways.

Using formula (1), and noting $\left|A\right|=\left(\begin{array}{c}13\\ 8\end{array}\right)·\left(\begin{array}{c}39\\ 18\end{array}\right)·\left(\begin{array}{l}26\\ 13\end{array}\right)·\left(\begin{array}{l}26\\ 13\end{array}\right)$(basic principle of counting)$P\left(A\right)=\frac{\left(\begin{array}{c}13\\ 8\end{array}\right)·\left(\begin{array}{c}39\\ 18\end{array}\right)·\left(\begin{array}{c}26\\ 13\end{array}\right)·\left(\begin{array}{c}26\\ 13\end{array}\right)}{\left|S\right|}$

Step 4: Count the number of elements in  

Count the number of elements in $AB$

Again, the choice of cards for North and South can be made. $\left(\begin{array}{c}13\\ 8\end{array}\right)\times \left(\begin{array}{c}39\\ 18\end{array}\right)\times \left(\begin{array}{l}26\\ 13\end{array}\right)$results. Choose whatever choices are made when selecting the cards for the North and South. $3$out of 5 spades and $10$out of$21$non spade cards for East, in $\left(\begin{array}{l}5\\ 3\end{array}\right)\left(\begin{array}{l}21\\ 10\end{array}\right)$, the remaining cards got to West.

Using formula (1), and noting $\left|AB\right|=\left(\begin{array}{c}13\\ 8\end{array}\right)\times \left(\begin{array}{c}39\\ 18\end{array}\right)\times \left(\begin{array}{l}26\\ 13\end{array}\right)\times \left(\begin{array}{l}5\\ 3\end{array}\right)\times \left(\begin{array}{l}21\\ 10\end{array}\right)$(basic principle$P\left(AB\right)=\frac{\left(\begin{array}{c}13\\ 8\end{array}\right)\times \left(\begin{array}{c}39\\ 18\end{array}\right)\times \left(\begin{array}{c}26\\ 13\end{array}\right)\times \left(\begin{array}{l}5\\ 3\end{array}\right)\times \left(\begin{array}{c}21\\ 10\end{array}\right)}{\left|S\right|}$

Find P(B|A)

$P(B\mid A)=\frac{\frac{\left(\begin{array}{c}13\\ 8\end{array}\right)\times \left(\begin{array}{c}39\\ 18\end{array}\right)\times \left(\begin{array}{c}26\\ 13\end{array}\right)\times \left(\begin{array}{c}5\\ 3\end{array}\right)\times \left(\begin{array}{c}21\\ 10\end{array}\right)}{\left|S\right|}}{\frac{\left(\begin{array}{c}13\\ 8\end{array}\right)\times \left(\begin{array}{l}39\\ 18\end{array}\right)\times \left(\begin{array}{c}26\\ 13\end{array}\right)\times \left(\begin{array}{c}26\\ 13\end{array}\right)}{\left|S\right|}}$

$\Rightarrow P(B\mid A)=\frac{\left(\begin{array}{l}5\\ 3\end{array}\right)\left(\begin{array}{l}21\\ 10\end{array}\right)}{\left(\begin{array}{c}26\\ 13\end{array}\right)}\approx 0.339$