Explanation:
We can approach this problem by considering the different cases in which we choose two counters of the same color, and then adding up their probabilities.
Case 1: Two grey counters are chosen. The probability of picking a grey counter on the first draw is 6/10, and the probability of picking another grey counter on the second draw (since we returned the first one) is also 6/10. Therefore, the probability of this case is (6/10) x (6/10) = 36/100.
Case 2: Two purple counters are chosen. The probability of picking a purple counter on the first draw is 1/10, and the probability of picking another purple counter on the second draw (since we returned the first one) is also 1/10. Therefore, the probability of this case is (1/10) x (1/10) = 1/100.
Case 3: Two white counters are chosen. The probability of picking a white counter on the first draw is 3/10, and the probability of picking another white counter on the second draw (since we returned the first one) is also 3/10. Therefore, the probability of this case is (3/10) x (3/10) = 9/100.
Since these three cases are mutually exclusive (i.e., they cannot happen at the same time), we can add up their probabilities to get the overall probability of choosing two counters of the same color:
36/100 + 1/100 + 9/100 = 46/100
Therefore, the probability of choosing two counters of the same color is 46/100, which can be simplified to 23/50.