Question

Bag $B_1$ contains 4 white and 2 black balls. Bag $B_2$ contains 3 white and 4 black balls. A bag is drawn at random and a ball is chosen at random from it. What is the probability that the ball drawn is white?

Answer 1

According to given in the question bag $B_1$ contains 4 white and 2 black balls and bag $B_2$ contains 3 white and 4 black balls but we have to find the probability that the drawn ball is white so, first of all, we have to choose one bag from the given to bags $B_1$ and $B_2$ and we have to find the probability that which bag is to choose from the given two bags. Now, as given in the question bag $B_1$ contains 4 white and 2 black balls but we have to choose one white ball from the 4 balls which will be our event and now same as we have to find the total outcomes from bag $B_1$ with the help of the formula given below:
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$…………………………(1)
Now, as given that in bag $B_2$ there are 3 white and 4 black balls but we have to find the probability that a white ball is drawn so first of all we have to one white ball from the 3 white balls which will be our event and now same as we have to find the total outcomes from bag $B_1$ with the help of the formula given above and now to find the probability we have to add both of the probability that a white ball is drawn from one of them.

Complete step by step answer:
Given,
Bag $B_1$, contains 4 white and 2 black balls
Bag $B_2$, contains 3 white and 4 black balls
Step 1: First of all we have to choose one bag from the given two bags as $B_1$ and $B_2$ hence to chose a bag is our is an event and the total number of bags is the total outcomes to find the probability
Probability to chose one bag from the given two bags ($B_1$ and $B_2$) = $\dfrac{1}{2}$
Step 2: Now, we have to find the probability that a white ball is drawn from the bag $B_1$. For this, as we know bag $B_1$ contains 4 white and 2 black balls but we have to choose one white ball from the 4 white balls which will be our event and now same as we have to find the total outcomes from bag $B_1$ to find the probability with the help of the formula (1) as mentioned in the solution hint.
Number of event to chose a white ball from the 4 white ball = $\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}$
Hence, on solving,
$= \dfrac{{4 \times 3!}}{{3!}} \\\ = 4 \\\$
Number of event to chose a white ball from the 4 white = 4
Now, same as total number of outcomes = $\dfrac{{6!}}{{1!\left( {6 - 1} \right)!}}$
Hence, on solving
$= \dfrac{{6 \times 5!}}{{5!}} \\\ = 6 \\\$
Hence the probability that the ball drawn is white from bag $B_1$ = $\dfrac{4}{6}$
Step 3: Hence, the probability that one white ball is drawn at random from the bag $B_1$ is:
$= \dfrac{1}{2} \times \dfrac{4}{6}$
On solving,
$= \dfrac{1}{3}$
The probability that one white ball is drawn at random from the bag $B_1$ is white $= \dfrac{1}{3}$
Step 4: Now, we have to find the probability that a white ball is drawn from the bag $B_2$. For this, as we know that bag $B_2$ contains 3 white and 4 black balls but we have to choose one white ball from the white balls which will be our event and now same as we have to find the total outcomes from the bag $B_2$ to find the probability with the help of the formula (1) as mentioned in the solution hint.
Number of event to chose a white ball from the 3 white ball =$\dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}$
On solving,
$= \dfrac{{3 \times 2!}}{{2!}} \\\ = 3 \\\$
Number of event to chose a white ball from the 3 white ball = 3
Now, same as total number of outcomes$= \dfrac{{7!}}{{7!\left( {7 - 1} \right)!}}$
On solving,
$= \dfrac{{7 \times 6!}}{{6!}} \\\ = 7 \\\$
Hence, the probability that the ball drawn is white from bag $B_2$ $= \dfrac{1}{2} \times \dfrac{3}{7}$
$= \dfrac{3}{{14}}$
Step 5: Now, to find the probability that a white ball is drawn from bag $B_1$ or $B_2$ we have to add the probability that the ball is drawn is white from bag $B_1$ with the probability that the ball is drawn is white from bag $B_{2}$.
Hence,
Required Probability $= \dfrac{1}{3} + \dfrac{3}{{14}}$
On solving,
$= \dfrac{{14 + 9}}{{42}} \\\ = \dfrac{{23}}{{42}} \\\$

Hence, probability that the ball is drawn is white from the bag $B_1$ which contains 4 white and 2 black balls, and the Bag $B_2$ which contains 3 white and 4 black balls is $= \dfrac{{23}}{{42}}$.

Note:
It is necessary that we find the probability for the bags $B_1$ and $B_2$ because as mentioned in the question that ball can be drawn from any one of the bags.
To chose the ball from the given number of balls we have to use the required formula to chose an object or thing and the formula is ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
It is required to add the probabilities of a white ball is drawn from bag $B_1$ and a white ball is drawn from bag $B_2$ because we don’t know from which bag a white ball is drawn from.