There are bags B1,B2 and B3. The bag B1 contains 5 red and 5 green balls, B2 contains 3 red 5 green balls, and B3 contains 5 red and 3 green balls, Bags B1,B2 and B3 have probabilities 310,310and 410respectively of being chosen. A bag is selected at random and a balls is chosen at random from the bag. Then which of the following options is/are correct?

Bi= Event for choosing bag Bi,i=1,2,3·G= Event for choosing a green ball. P(G)=PB1PGB1+PB2PGB2+PB3PGB3=310×510+310×58+410×38=320+316+320=3980 Option 1 correct PB3G=PB3∩GP(G)=410×383980=413 option 2 incorrect PGB3=38 option 3 is correct PB3∩G=PB3·PGR=410×38=320 option 4 incorrect

There are bags B1,B2 and B3. The bag B1 contains 5 red and 5 green balls, B2 contains 3 red 5 green balls, and B3 contains 5 red and 3 green balls, Bags B1,B2 and B3 have probabilities 310,310and 410respectively of being chosen. A bag is selected at random and a balls is chosen at random from the bag. Then which of the following options is/are correct?

Bi= Event for choosing bag Bi,i=1,2,3·G= Event for choosing a green ball. P(G)=PB1PGB1+PB2PGB2+PB3PGB3=310×510+310×58+410×38=320+316+320=3980 Option 1 correct PB3G=PB3∩GP(G)=410×383980=413 option 2 incorrect PGB3=38 option 3 is correct PB3∩G=PB3·PGR=410×38=320 option 4 incorrect

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There are bags $B_{1}, B_{2}$ and $B_{3}$ . The bag $B_{1}$ contains 5 red and 5 green balls, $B_{2}$ contains 3 red 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 balls is chosen at random from the bag. Then which of the following options is/are correct?

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Probability that the chosen balls 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 chosen ball is green, given that the selected bag is $B_{3}$ equals $\frac{3}{8}$

Probability that the selected bag is $B_{3}$ and the chosen ball is green equals $\frac{3}{10}$

answer is A, C.

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Detailed Solution

$B_{i} = Event for choosing bag B_{i}, i = 1, 2, 3 \cdot G = Event for choosing a green ball.$

$\begin{array}{l} P (G) = P (B_{1}) P (\frac{G}{B_{1}}) + P (B_{2}) P (\frac{G}{B_{2}}) + P (B_{3}) P (\frac{G}{B_{3}}) \\ = \frac{3}{10} \times \frac{5}{10} + \frac{3}{10} \times \frac{5}{8} + \frac{4}{10} \times \frac{3}{8} = \frac{3}{20} + \frac{3}{16} + \frac{3}{20} = \frac{39}{80} Option 1 correct \end{array}$

$\begin{array}{l} P (\frac{B_{3}}{G}) = \frac{P (B_{3} \cap G)}{P (G)} = \frac{\frac{4}{10} \times \frac{3}{8}}{\frac{39}{80}} = \frac{4}{13} option 2 incorrect \\ P (\frac{G}{B_{3}}) = \frac{3}{8} option 3 is correct \\ P (B_{3} \cap G) = P (B_{3}) \cdot P (\frac{G}{R}) = \frac{4}{10} \times \frac{3}{8} = \frac{3}{20} option 4 incorrect \end{array}$