# A fair die is rolled. If face 1 turns up, a ball is drawn from Bag A. If face 2 or 3 turns up, a ball is drawn from Bag B. If face 4 or 5 or 6 turns up, a ball is drawn from Bag C. Bag A contains 3 red and 2 white balls, Bag B contains 3 red and 4 white balls and Bag C contains 4 red and 5 white balls. The die is rolled, a Bag is picked up and a ball is drawn. If the drawn ball is red; what is the probability that it is drawn from Bag B?

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**Hint**: To solve this question, we need to use the basic theory related to the probability. As we know, to calculate the probability of an event, we take the number of successes, divided by the number of possible outcomes.

The probability of event A happening is:

P(A) = $ $ $ \dfrac{{{\text{n}}\left( {\text{A}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} $

**:**

__Complete step-by-step answer__As given, A fair die is rolled and Bag A contains 3 red and 2 white balls, Bag B contains 3 red and 4 white balls and Bag C contains 4 red and 5 white balls.

Now,

Probability that ball is drawn from bag A = $ {\text{P}}\left( {{{\text{E}}_{\text{1}}}} \right) $ = $ \dfrac{1}{6} $

Probability that ball is drawn from bag B = $ {\text{P}}\left( {{{\text{E}}_2}} \right) $ = $ \dfrac{2}{6} $

Probability that ball is drawn from bag C = $ {\text{P}}\left( {{{\text{E}}_3}} \right) $ = $ \dfrac{3}{6} $

let P(A) be the probability of drawing a red ball

Probability that it is drawn from bag A = $ {\text{P}}\left( {{\text{A|}}{{\text{E}}_{\text{1}}}} \right) $ = $ \dfrac{3}{5} $

Probability that it is drawn from bag B = $ {\text{P}}\left( {{\text{A|}}{{\text{E}}_2}} \right) $ = $ \dfrac{3}{7} $

Probability that it is drawn from bag C = $ {\text{P}}\left( {{\text{A|}}{{\text{E}}_3}} \right) $ = $ \dfrac{4}{9} $

Now we have to find $ {\text{P}}\left( {{{\text{E}}_{\text{2}}}{\text{|A}}} \right) $ that is it is drawn from bag B

Using Baye's theorem

$ {\text{P}}\left( {{{\text{E}}_{\text{2}}}{\text{|A}}} \right) $ = $ \dfrac{{{\text{P}}\left( {{{\text{E}}_{\text{2}}}} \right){\text{P}}\left( {{\text{A|}}{{\text{E}}_{\text{2}}}} \right)}}{{{\text{P}}\left( {{{\text{E}}_1}} \right){\text{P}}\left( {{\text{A|}}{{\text{E}}_1}} \right) + {\text{P}}\left( {{{\text{E}}_{\text{2}}}} \right){\text{P}}\left( {{\text{A|}}{{\text{E}}_{\text{2}}}} \right) + {\text{P}}\left( {{{\text{E}}_3}} \right){\text{P}}\left( {{\text{A|}}{{\text{E}}_3}} \right)}} $

$ {\text{P}}\left( {{{\text{E}}_{\text{2}}}{\text{|A}}} \right) $ = $ \dfrac{{\dfrac{1}{3} \times \dfrac{3}{7}}}{{\dfrac{1}{6} \times \dfrac{3}{5} + \dfrac{1}{3} \times \dfrac{3}{7} + \dfrac{1}{2} \times \dfrac{4}{9}}} $

$ {\text{P}}\left( {{{\text{E}}_{\text{2}}}{\text{|A}}} \right) $ = $ \dfrac{{\dfrac{1}{7}}}{{\dfrac{1}{{10}} + \dfrac{1}{7} + \dfrac{2}{9}}} $

$ {\text{P}}\left( {{{\text{E}}_{\text{2}}}{\text{|A}}} \right) $ = $ \dfrac{{\dfrac{1}{7}}}{{\dfrac{{293}}{{630}}}} $

$ {\text{P}}\left( {{{\text{E}}_{\text{2}}}{\text{|A}}} \right) $ = $ \dfrac{{90}}{{293}} $

$ {\text{P}}\left( {{{\text{E}}_{\text{2}}}{\text{|A}}} \right) $ = 0.307

Therefore, the probability that it is drawn from Bag B is $ \dfrac{{90}}{{293}} $ or 0.307.

**Note**: Probability is a number between 0 and 1. We multiply our final answer by 100 to get a “percent”. We borrowed that concept from the French - “per cent” - which means per 100. Something with 0 or 0% probability will never happen. Something with 1 or 100% percent probability will always happen. Something with 0.5 or 50% probability will happen half the time.

If your final answer is smaller than 0 or bigger than 1, then you made a mistake. Everything happens between 0 and 1.