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?
- Probability that the chosen ball is green, given that the selected bag is $B_3$, equals $\frac{3}{8}$
- Probability that the chosen ball is green equals $\frac{39}{80}$
- Probability that the selected bag is $B_3$, given that the chosen ball is green, equals $\frac{5}{13}$
- Probability that the selected bag is $B_3$ and the chosen ball is green equals $\frac{3}{10}$
The Correct Option is B
Solution and Explanation
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Concepts Used:
Multiplication Theorem on Probability
In accordance with the multiplication rule of probability, the probability of happening of both the events A and B is equal to the product of the probability of B occurring and the conditional probability that event A happens given that event B occurs.
Let's assume, If A and B are dependent events, then the probability of both events occurring at the same time is given by:
\(P(A\cap B) = P(B).P(A|B)\)
Let's assume, If A and B are two independent events in an experiment, then the probability of both events occurring at the same time is given by:
\(P(A \cap B) = P(A).P(B)\)
Read More: Multiplication Theorem on Probability