Question

One bag contains 3 white balls, 7 red balls and 15 black balls. Another bag contains 10 white balls, 6 red balls and 9 black balls. One ball is taken from each bag. What is the probability that both the balls will be of the same colour?\[\]

Answer 1

Hint: We get three cases to take two balls of the same colour. . We can select a white ball from the first and second bag. We find its probability $P\left( W \right)$ taking the total number of balls as sample size and number of white balls as size of the event. We can do the same for black and red balls and get the probabilities $P\left( R \right),P\left( B \right)$. We add up the three probabilities to get the result. \[\]

Complete step-by-step answer:
We know from definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring and $n\left( S \right)$ is the size of the sample space then the probability of the event $A$ occurring is $\dfrac{n\left( A \right)}{n\left( S \right)}$.
It is given in the question that the first bag contains 3 white balls, 7 red balls and 15 black balls. So the total number of balls in the first bag is $3+7+15=25$. So 25 is the size of sample space for first bag.\[\]
It is also given that the second bag contains 10 white balls, 6 red balls and 9 black balls So the total number of balls in the first bag is $10+6+9=25$. So 25 is the size of sample space for second bag.\[\]
We have to select one ball from each bag of the same colour. Selecting a ball from one bag does not affect the selection of the ball from another bag. \[\]
Case-1: We select a white ball from the first bag and then the second bag out of balls . There are 3 white balls in the first bag and 10 in the second. So the probability of such selection is
\[P\left( W \right)=\dfrac{3}{25}\times \dfrac{10}{25}\]
Case-2: We select a red ball from the first bag and then the second bag. There are 7 red balls in the first bag and 6 in the second. So the probability of such selection is
\[P\left( R \right)=\dfrac{7}{25}\times \dfrac{6}{25}\]

Case-3: We select a black ball from the first bag and then the second bag. There are 15 black balls in the first bag and 9 in the second. So the probability of such selection is
\[P\left( B \right)=\dfrac{15}{25}\times \dfrac{9}{25}\]
So the three cases cannot occur at the same time. So the probability of selecting two balls of same colour from different bag is
\[P=P\left( W \right)+P\left( R \right)+P\left( B \right)=\dfrac{3}{25}\times \dfrac{10}{25}+\dfrac{7}{25}\times \dfrac{6}{25}+\dfrac{15}{25}\times \dfrac{9}{25}=\dfrac{207}{625}\]

Note: We need to be careful that the balls are identical and we did not use the combination formula. If they would we have been unique, we would have used the formula $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ for selection of $r$ balls from $n$ unique balls.