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Question

Mathematics Question on Multiplication Theorem on Probability

There are three bags B1,B2B_1, B_2 and B3B_3. The bag B1B_1 contains 5 red and 5 green balls, B2B_2 contains 3 red and 5 green balls, and B3B_3 contains 5 red and 3 green balls. Bags B1,B2B_1, B_2 and B3B_3 have probabilities 310,310\frac{3}{10}, \frac{3}{10} and 410\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?

A

Probability that the chosen ball is green, given that the selected bag is B3B_3, equals 38\frac{3}{8}

B

Probability that the chosen ball is green equals 3980\frac{39}{80}

C

Probability that the selected bag is B3B_3, given that the chosen ball is green, equals 513\frac{5}{13}

D

Probability that the selected bag is B3B_3 and the chosen ball is green equals 310\frac{3}{10}

Answer

Probability that the chosen ball is green equals 3980\frac{39}{80}

Explanation

Solution

P(B1)=310P(B2)=310P(B3)=410P\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 B3B_3 and the chosen ball is green =P(B3)×P(GB3)=P\left(B_{3}\right)\times P\left(\frac{G}{B_{3}}\right) =410×38=320=\frac{4}{10}\times\frac{3}{8}=\frac{3}{20} (B) Probability that the selected bag isB3 B_3 given that the chosen ball is green P(B3G)P\left(\frac{B_{3}}{G}\right) P(B3G)=P(GB3)P(B3)P(GB1)P(B1)+P(GB2)P(B2)+P(GB3)P(B3)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)} =410×38310×510+310×58+410×38=413=\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 B3B_3 P(GB3)=38P\left(\frac{G}{B_{3}}\right)=\frac{3}{8} (D) Probability that the chosen ball is green P(G)=P(B1)P(GB1)+P(B2)P(GB2)+P(B3)P(GB3)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) =310×510+310×58+410×38=\frac{3}{10}\times\frac{5}{10}+\frac{3}{10}\times\frac{5}{8}+\frac{4}{10}\times\frac{3}{8} =3980=\frac{39}{80}