Question

Contents of the two urns is as given in this table. A fair die is tossed. If the face 1,2,4 or 5 comes, a marble is drawn from the urn A other wise a marble is chosen from the urn B.
$\begin{array}{cccc}Urn& RedMarbles& Whitemarbles& Bluemarbles\\ A& 5& 3& 8\\ B& 3& 5& 0\end{array}$
Let $E_1$: Denote the event that a red marble is chosen
$E_2$: Denote the event that a white marble is chosen
$E_3$: Denote the event that a blue is chosen

• A. The event $E_1,E_2$ and $E_3$ are equi probable
• B. $P\left(E_1\right),P\left(E_2\right),P\left(E_3\right)$ are in A.P.
• C. If the marble drawn is red, the probability that it came from the urn A is $\left(\frac{1}{2}\right)$
• D. If the marble drawn is white, the probability that the face 5 appeared on the die is $\left(\frac{3}{32}\right)$

Answer 1

Answer: (A), (B), (D)

The correct options are
A The event $E_1,E_2$ and $E_3$ are equi probable
B $P\left(E_1\right),P\left(E_2\right),P\left(E_3\right)$ are in A.P.
D If the marble drawn is white, the probability that the face 5 appeared on the die is $\left(\frac{3}{32}\right)$

$\begin{array}{cccc}Urn& RedMarbles& Whitemarbles& Bluemarbles\\ A& 5& 3& 8\\ B& 3& 5& 0\end{array}$
$P\left(E_1\right)=P\left(R\right)=\left(\frac{2}{3}\right)\left(\frac{5}{16}\right)+\left(\frac{1}{3}\right)\left(\frac{3}{8}\right)=\frac{10}{48}+\frac{6}{48}=\frac{1}{3}P\left(E_2\right)=P\left(W\right)=\left(\frac{2}{3}\right)\left(\frac{3}{16}\right)+\left(\frac{1}{3}\right)\left(\frac{5}{8}\right)=\frac{6}{48}+\frac{10}{48}=\frac{1}{3}P\left(E_3\right)=P\left(B\right)=\left(\frac{2}{3}\right)\left(\frac{8}{16}\right)=\frac{1}{3}$
(C) Let A : event that urn A is chose
$P\left(\frac{A}{R}\right)=\frac{P(A\cap R)}{P\left(R\right)}=\frac{\left(\frac{2}{3}\right)\left(\frac{5}{16}\right)}{\frac{1}{3}}=\left(\frac{10}{48}\right)\left(3\right)=\frac{5}{8}\Rightarrow \left(C\right)$ isincorrect.
$\left(D\right)P\left(\frac{A}{W}\right)=\frac{P(A\cap W)}{P\left(W\right)}=\frac{\left(\frac{2}{3}\right)\left(\frac{3}{16}\right)}{\frac{1}{3}}=\left(\frac{6}{48}\right)\left(3\right)=\frac{3}{8}P\left(\frac{facefive}{W}\right)=\left(\frac{3}{8}\right)\left(\frac{1}{4}\right)=\frac{3}{32}\Rightarrow \left(D\right)$ is correct]