Question:
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?
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?
Updated On: Aug 1, 2022
- 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 $\frac{3}{10}$
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The Correct Option is B
Solution and Explanation
$P\left(B_{1}\right)=\frac{3}{10} \,P\left(B_{2}\right)=\frac{3}{10}\, P\left(B_{3}\right)=\frac{4}{10}$
(A) Probability that selected bag is $B_3$ and the chosen ball is green
$=P\left(B_{3}\right)\times P\left(\frac{G}{B_{3}}\right)$
$=\frac{4}{10}\times\frac{3}{8}=\frac{3}{20}$
(B) Probability that the selected bag is$ B_3$ given that the chosen ball is green $P\left(\frac{B_{3}}{G}\right)$
$P\left(\frac{B_{3}}{G}\right)=\frac{P\left(\frac{G}{B_{3}}\right)P\left(B_{3}\right)}{P\left(\frac{G}{B_{1}}\right)P\left(B_{1}\right)+P\left(\frac{G}{B_{2}}\right)P\left(B_{2}\right)+P\left(\frac{G}{B_{3}}\right)P\left(B_{3}\right)}$
$=\frac{\frac{4}{10}\times\frac{3}{8}}{\frac{3}{10}\times\frac{5}{10}+\frac{3}{10}\times\frac{5}{8}+\frac{4}{10}\times\frac{3}{8}}=\frac{4}{13}$
(C) Probability that the chosen ball is green, given that the selected bag is $B_3$
$P\left(\frac{G}{B_{3}}\right)=\frac{3}{8}$
(D) Probability that the chosen ball is green
$P\left(G\right)=P\left(B_{1}\right)P\left(\frac{G}{B_{1}}\right)+P\left(B_{2}\right)P\left(\frac{G}{B_{2}}\right)+P\left(B_{3}\right)P\left(\frac{G}{B_{3}}\right)$
$=\frac{3}{10}\times\frac{5}{10}+\frac{3}{10}\times\frac{5}{8}+\frac{4}{10}\times\frac{3}{8}$
$=\frac{39}{80}$
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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:
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