Question

Two balls are drawn from a bag containing $5$ white and $7$ black balls at random. What is the probability that they would be of different colours?

Answer 1

For solving this problem we need to have a clear understanding of probability along with permutations and combinations. Using the predefined concepts and formula, and calculating the events which are supposed to occur along with the total events, we can easily solve the question.

Complete step-by-step answer:
Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true. The probability of an event is a number between $0$ and $1$ , where, roughly speaking, $0$ indicates impossibility of the event and $1$ indicates certainty. The higher the probability of an event, the more likely it is that the event will occur. A simple example is the tossing of a fair (unbiased) coin. Since the coin is fair, the two outcomes ("heads" and "tails") are both equally probable; the probability of "heads" equals the probability of "tails"; and since no other outcomes are possible, the probability of either "heads" or "tails" is $\dfrac{1}{2}$ (which could also be written as $0.5$ or $50$ ).
Now according to the given problem, two balls are drawn at random from a bag containing $5$ white and $7$ black balls. We want to determine the probability that they are of different colours.
This can happen when the first ball is white and the second black or if the first ball is black and the second white. The probability that the first ball is white is $\dfrac{5}{12}$ . Then the probability that the second ball is black is $\dfrac{7}{11}$ .
⇒ The probability that the first ball is white and the second black is $\dfrac{5}{12}\times \dfrac{7}{11}=\dfrac{35}{132}$ .
The probability that the first ball is black is $\dfrac{7}{12}$ . Then the probability that the second ball is white is $\dfrac{5}{11}$ .
⇒ The probability that the first ball is black and the second red is $\dfrac{7}{12}\times \dfrac{5}{11}=\dfrac{35}{132}$ .
⇒ The probability that the balls are of different colours is $\dfrac{35}{132}+\dfrac{35}{132}=\dfrac{35}{66}$ .
Hence, the answer to the given problem is $\dfrac{35}{66}$ .

Note: These questions are pretty easy to solve but one needs to be very careful while performing the calculations. The common mistake committed in this problem, is that one may replace the ball after drawing. Hence, these minute errors should be kept in mind to get the correct answer.