Question

A bag ‘A’ contains \[2\] white and \[3\] red balls and bag ‘B’ contains \[4\] white and \[5\] red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag ‘B’ was
A) \[\dfrac{5}{{14}}\]
B) \[\dfrac{5}{{16}}\]
C) \[\dfrac{5}{{18}}\]
D) \[\dfrac{{25}}{{52}}\]

Answer 1

Hint: Probability means possibility which deals with the occurrence of a random event or to predict how likely events are to happen. The value is expressed from zero to one. Here, we will use the Bayes’ theorem formula i.e.\[P(\dfrac{A}{B}) = \dfrac{{P(A)P(\dfrac{A}{B})}}{{P(B)}}\], \[P(B) \ne 0\] where A and B are the events.

Complete step by step answer:
Let the probability of choosing bag A or the event of selecting the bag A is denoted by ‘A’, the probability of choosing the bag B or the event of selecting the bag B is denoted by ‘B’ and let the probability of red ball drawn is red or the event of selecting the red ball is denoted by ‘R’.
Given that, Bag A has \[2\] white and \[3\] red balls and the total number of balls \[ = 2 + 3 = 5\] balls.
Then, the probability of R from A is = P (R/A) = \[\dfrac{3}{5}\]
And, for bag B has \[4\] white and \[5\] red balls and the total number of balls \[ = 4 + 5 = 9\] balls.
Then, the probability of R from B is = P (R/B) = \[\dfrac{5}{9}\]
We have to find the probability that drawing a ball from bag B is red = P (B/R).
Since, both the bags are equally likely to be selected, so we will have,
The probability of event A is P (A) = the probability of event B is P (B) =\[\dfrac{1}{2}\].
From Bayes' theorem, we get,
\[P(\dfrac{B}{R}) = \dfrac{{P(B)P(\dfrac{R}{B})}}{{P(R)}}\]
\[ = \dfrac{{P(B)P(\dfrac{R}{B})}}{{P(\dfrac{R}{A})P(A) + P(\dfrac{R}{B})P(B)}}\]
Substituting the values in the above expression, we will get,
\[ = \dfrac{{(\dfrac{1}{2})(\dfrac{5}{9})}}{{(\dfrac{3}{5})(\dfrac{1}{2}) + (\dfrac{5}{9})(\dfrac{1}{2})}}\]
Taking\[\dfrac{1}{2}\]common from both numerator and denominator, we will get,
\[ = \dfrac{{(\dfrac{1}{2})(\dfrac{5}{9})}}{{\dfrac{1}{2}(\dfrac{3}{5} + \dfrac{5}{9})}}\]
\[ = \dfrac{{\dfrac{5}{9}}}{{\dfrac{3}{5} + \dfrac{5}{9}}}\]
Taking LCM (\[5\],\[9\]) =\[45\], we will get,
\[ = \dfrac{{\dfrac{5}{9}}}{{\dfrac{{27 + 25}}{{45}}}}\]
\[ = \dfrac{{\dfrac{5}{9}}}{{\dfrac{{52}}{{45}}}}\]
\[ = \dfrac{5}{9} \div \dfrac{{52}}{{45}}\]
\[ = \dfrac{5}{9} \times \dfrac{{45}}{{52}}\]
\[ = 5 \times \dfrac{5}{{52}}\]
\[ = \dfrac{{25}}{{52}}\]

Hence, the probability that the red ball is drawn from bag B is \[\dfrac{{25}}{{52}}\].

Note:
Bayes’ theorem describes the probability of occurrence of an event related to any condition. It is also considered for the case of conditional probability. Bayes theorem is also known as the formula for the probability of “causes”. The formula to find the probability i.e. Probability of events to happen $P(E) = \dfrac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}$. Sometimes students get mistaken for a favourable outcome with a desirable outcome.