Question

Bag 1 contains 3 black and 4 white balls and bag 2 contains 4 black and 3 white balls. A die is thrown. If it shows 1 or 3, then a ball is drawn from bag 1 and if some other number appears, a ball is drawn from bag 2. Find the probability that the drawn ball is black.

Answer 1

Hint: In this question, we are given 2 bags with different colours of balls and we have been asked to find the probability of drawing a black ball on the condition of throwing a dice. Son now we have two events- throwing a dice and drawing a ball. Find out the probability of both the events individually and then use total probability to find the desired answer.

Complete step-by-step answer:
We are given 2 bags. Bag 1 has 3 black and 4 white balls. Bag 2 has 4 black and 3 white balls. There are 2 events-
Event 1: A dice is thrown.
Event 2: A ball is drawn.
Since we have been asked to find the probability of the $2^{nd}$ event, we will have to find the total probability.
If a dice is thrown and 1 or 3 shows up, the ball is drawn from bag 1. On the other hand if 2,4,5 or 6 shows up, then the ball is drawn from bag 2.
Let ${E_1}$ be the event that 1 or 3 shows up on the dice.
$\Rightarrow$$P({E_1}) = \dfrac{2}{6} = \dfrac{1}{3}$
Let ${E_2}$ be the event that 2,4,5 or 6 shows up on the dice.
$\Rightarrow$$P({E_2}) = \dfrac{4}{6} = \dfrac{2}{3}$
Let $A$ be the event that black ball is drawn. Let us find the probability of drawing a black ball when 1 or 3 shows up on the dice. It is denoted by $P(A/{E_1})$.
$P(A/{E_1}) = \dfrac{3}{7}$ (If 1 or 3 shows up, ball is to be drawn from bag 1 and there are 3 black balls in bag 1)
Let us find the probability of drawing a black ball when 2,4,5 or 6 shows up on the dice. It is denoted by $P(A/{E_2})$.
$P(A/{E_2}) = \dfrac{4}{7}$ (In this case, ball is to drawn from bag 2 and there are 4 black balls in this bag)
Now we will find $P(A)$ using the total probability theorem.
$P(A) = P({E_1})P(A/{E_1}) + P({E_2})P(A/{E_2})$
Putting all the values in the above equation,
$\Rightarrow$$P(A) = \dfrac{1}{3} \times \dfrac{3}{7} + \dfrac{2}{3} \times \dfrac{4}{7}$
Solving,
$\Rightarrow$$P(A) = \dfrac{3}{{21}} + \dfrac{8}{{21}} = \dfrac{{11}}{{21}}$

Therefore, the probability that the drawn ball is black $ = \dfrac{{11}}{{21}}$

Note: Students often get confused between total probability and bayes’ theorem. The easy way to tell which one of the two will be applicable is to see which probability is to be found out. In every question, there will be 2 events. If the probability of the $1^{st}$ event is asked, use bayes’ theorem. If the probability of the $2^{nd}$ event is asked, use the total probability theorem.