Question

A bag contains 4 red and 5 black balls, a second bag contains 3 red and 7 black balls. One ball is drawn at random from each bag. Find the probability that the ball is of the same color.

Answer 1

Hint: In this question, first of all, calculate the probability of drawing the red and black balls from both the bags individually. Now to get the probability that the balls are of the same color, add the probability of drawing the red and drawing the black ball.

Complete step-by-step answer:
Here we are given that a bag contains 4 red and 5 black balls and a second bag contains 3 red and 7 black balls. One ball is drawn at random from each bag, we have to find the probability that the balls are of the same color. Before proceeding with this question, let us first understand what probability is. Probability is a measure of the likelihood of an event to occur. We can predict the chance of an event to occur by using it. Probability means possibility. Also, the probability of the event to happen $=\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
Now, we know that in the first bag, there are 4 red balls and 5 black balls. So, we get,
The probability of drawing a red ball from bag 1, $P\left( {{R}_{1}} \right)=\dfrac{\text{Number of red balls in bag 1}}{\text{Total balls in bag 1}}$
So, we get, $P\left( {{R}_{1}} \right)=\dfrac{4}{9}$
The probability of drawing a black ball from bag 1, $P\left( {{B}_{1}} \right)=\dfrac{\text{Number of black balls in bag 1}}{\text{Total balls in bag 1}}$
So, we get, $P\left( {{B}_{1}} \right)=\dfrac{5}{9}$
Now, we also know that in the second bag, there are 3 red balls and 7 black balls. So, we get,
The probability of drawing a red ball from bag 2, $P\left( {{R}_{2}} \right)=\dfrac{\text{Number of red balls in bag 2}}{\text{Total balls in bag 2}}$
So, we get, $P\left( {{R}_{2}} \right)=\dfrac{3}{10}$
The probability of drawing a black ball from bag 2, $P\left( {{B}_{2}} \right)=\dfrac{\text{Number of black balls in bag 2}}{\text{Total balls in bag 2}}$
So, we get, $P\left( {{B}_{2}} \right)=\dfrac{7}{10}$
So, the probability of drawing a red ball from both the bags 1 and 2, $\left( P\left( R \right) \right)=P\left( {{R}_{1}} \right).P\left( {{R}_{2}} \right)$
By substituting the value of $P\left( {{R}_{1}} \right)=\dfrac{4}{9}$ and $P\left( {{R}_{2}} \right)=\dfrac{3}{10}$. So, we get,
$P\left( R \right)=\dfrac{4}{9}.\dfrac{3}{10}=\dfrac{12}{90}$
Also, the probability of drawing a black ball from both the bags 1 and 2, $\left( P\left( B \right) \right)=P\left( {{B}_{1}} \right).P\left( {{B}_{2}} \right)$
By substituting the value of $P\left( {{B}_{1}} \right)=\dfrac{5}{9}$ and $P\left( {{B}_{2}} \right)=\dfrac{7}{10}$. So, we get,
$P\left( B \right)=\dfrac{5}{9}.\dfrac{7}{10}=\dfrac{35}{90}$
So, the probability of drawing the ball of the same color P(T) = P(R) + P(B)
By substituting the value of $P\left( R \right)=\dfrac{12}{90}$ and $P\left( B \right)=\dfrac{35}{90}$, we get,
$P\left( T \right)=\dfrac{12}{90}+\dfrac{35}{90}$
$P\left( T \right)=\dfrac{47}{90}$
So, we get the probability of drawing a ball of the same color as $\dfrac{47}{90}$.

Note: In this question, many students make this mistake of adding probabilities of two events instead of multiplying and multiplying instead of adding. So, they must take care of that. Generally, whenever there is AND between probabilities of two events, we multiply them whereas if there is OR between probabilities of two events, we add them. So, this concept should be there in mind.