An urn contains 5 red and 7 green balls. A ball is drawn at random and...
The Problem
An urn contains 5 red and 7 green balls. A ball is drawn at random and its colour is noted. The ball is placed back into the urn along with another ball of the same colour. We need to find the probability of getting a red ball in the next draw.
Analysis
To solve this problem, we need to consider two scenarios:
1. The first ball drawn is red.
2. The first ball drawn is green.
Scenario 1: The first ball drawn is red
In this case, the probability of drawing a red ball is 5/12 (since there are 5 red balls and a total of 12 balls in the urn). After drawing the first red ball, it is placed back in the urn along with another red ball. So the total number of balls in the urn remains the same (12).
Scenario 2: The first ball drawn is green
In this case, the probability of drawing a green ball is 7/12 (since there are 7 green balls and a total of 12 balls in the urn). After drawing the first green ball, it is placed back in the urn along with another green ball. So the total number of balls in the urn remains the same (12).
Calculating the probability
To calculate the probability of getting a red ball in the next draw, we need to consider both scenarios and add up the probabilities.
Probability of the first ball being red * Probability of getting a red ball in the next draw given the first ball was red:
(5/12) * (6/12) = 30/144
Probability of the first ball being green * Probability of getting a red ball in the next draw given the first ball was green:
(7/12) * (5/12) = 35/144
Adding up the probabilities:
(30/144) + (35/144) = 65/144
Simplifying the fraction:
65/144 = 13/28
Therefore, the probability of getting a red ball in the next draw is 13/28.
Conclusion
The correct answer is option 'A' (65/156).