There are four boxes, $A, B, C$ and $D$, containing marbles. A contains 1 red, 6 white and 3 black marbles; $B$ contains 6 red, 2 white and 2 black marbles; C contains 8 red, 1 white and 1 black marbles; and D contains 6 white and 4 black marbles. One of the boxes is selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from the box $A$ ?

Home » There are four boxes, and , containing marbles. A contains 1 red, 6 white and 3 black marbles; contains 6 red, 2 white and 2 black marbles; C contains 8 red, 1 white and 1 black marbles; and D contains 6 white and 4 black marbles. One of the boxes is selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from the box ?

There are four boxes, $A, B, C$ and $D$ , containing marbles. A contains 1 red, 6 white and 3 black marbles; $B$ contains 6 red, 2 white and 2 black marbles; C contains 8 red, 1 white and 1 black marbles; and D contains 6 white and 4 black marbles. One of the boxes is selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from the box $A$ ?

There are four boxes, $A, B, C$ and $D$ , containing marbles. A contains 1 red, 6 white and 3 black marbles; $B$ contains 6 red, 2 white and 2 black marbles; C contains 8 red, 1 white and 1 black marbles; and D contains 6 white and 4 black marbles. One of the boxes is selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from the box $A$ ?

Let $A:$ Ball drawn from bag $A$
B: Ball is drawn from bag B
$C:$ Ball is drawn from bag $C$
$D:$ Ball is drawn from bag $D$
BB: Black ball
WB : White ball
RB : Red ball
Assuming all boxes have an equal probability for picking i.e. $\frac{1}{4}$
We want to find $P(A \mid R B)$ , i.e. probability of selected red ball is from box $A$
$\begin{array}{l} P(A \mid R B)=\frac{P(A) \cdot P(R B \mid A)}{P(A) \cdot P(R B \mid A)+P(B) \cdot P(R B \mid B)+P(C) \cdot P(R B \mid C)+P(D) \cdot P(R B \mid D)} \\ =\frac{\left(\frac{1}{4}\right)\left(\frac{1}{10}\right)}{\left(\frac{1}{4}\right)\left(\frac{1}{10}\right)+\left(\frac{1}{4}\right)\left(\frac{6}{10}\right)+\left(\frac{1}{4}\right)\left(\frac{8}{10}\right)+\left(\frac{1}{4}\right)\left(\frac{0}{10}\right)} \\ =\frac{1}{15} \end{array}$
Therefore, the probability of selected red ball is from box $A$ is $\frac{1}{15}$