There are three bags $B_1, B_2$ and $B_3$. The bag $B_1$ contains $5$ red and $5$ green balls,$B_2$ contains $3$ red and $5$ green balls,and $B_3$ contains $5$ red and $3$ green balls. Bags $B_1, B_2$ and $B_3$ have probabilities $\frac{3}{10}, \frac{3}{10}$ and $\frac{4}{10}$ respectively of being chosen. $A$ bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
$(1)$ Probability that the selected bag is $B_3$ and the chosen ball is green equals $\frac{3}{20}$
$(2)$ Probability that the chosen ball is green equals $\frac{39}{80}$
$(3)$ Probability that the chosen ball is green,given that the selected bag is $B_3$,equals $\frac{3}{8}$
$(4)$ Probability that the selected bag is $B_3$,given that the chosen ball is green,equals $\frac{4}{13}$

  • A
    $1, 2$
  • B
    $1, 3$
  • C
    $2, 3$
  • D
    $3, 4$

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Let $U_1$ and $U_2$ be two urns such that $U_1$ contains $3$ white and $2$ red balls,and $U_2$ contains only $1$ white ball. $A$ fair coin is tossed. If it shows head,$1$ ball is drawn at random from $U_1$ and transferred to $U_2$. If it shows tail,$2$ balls are drawn at random from $U_1$ and transferred to $U_2$. Now,$1$ ball is drawn at random from $U_2$. Given that the ball drawn from $U_2$ is white,what is the probability that the coin showed head?

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