An urn contains n > 0 white balls and m > 0 black balls. Consider the experiment where two balls are drawn without replacement. What is the probability that the balls are of the same color? What is the probability of this event, if the balls are drawn with replacement? Which of these numbers is larger? (Explain your answer.)
To find the probability that the two balls drawn are of the same color without replacement, we need to consider two cases: either both balls are white or both balls are black.
Case 1: Both balls are white
The probability of drawing a white ball on the first draw is n / (n + m) because there are n white balls out of a total of (n + m) balls. After drawing a white ball, there will be (n - 1) white balls left and (m) black balls left. Therefore, the probability of drawing another white ball is (n - 1) / (n + m - 1). To calculate the probability of both events happening (drawing two white balls), we multiply these probabilities together:
P(Both balls are white) = (n / (n + m)) * ((n - 1) / (n + m - 1))
Case 2: Both balls are black
Similarly, the probability of drawing a black ball on the first draw is m / (n + m). After drawing a black ball, there will be (n) white balls left and (m - 1) black balls left. Therefore, the probability of drawing another black ball is (m - 1) / (n + m - 1). To calculate the probability of both events happening (drawing two black balls), we multiply these probabilities together:
P(Both balls are black) = (m / (n + m)) * ((m - 1) / (n + m - 1))
The probability that the balls are of the same color without replacement is the sum of the probabilities of both cases:
P(Same color without replacement) = P(Both balls are white) + P(Both balls are black)
To find the probability of the event if the balls are drawn with replacement, we assume that after each ball is drawn, it is put back into the urn before the next ball is drawn. In this case, the probability of drawing either a white or a black ball remains the same for each draw, and the events of drawing the balls are independent. Therefore, the probability of drawing two balls of the same color with replacement is the product of the probabilities of drawing either two white balls or two black balls:
P(Same color with replacement) = P(Both balls are white with replacement) + P(Both balls are black with replacement) = (n / (n + m)) * (n / (n + m)) + (m / (n + m)) * (m / (n + m))
To compare the probabilities, we can simplify the expressions and compare them directly. However, before simplification, we can make an observation:
Notice that in the expression P(Same color with replacement), the terms being multiplied (n / (n + m)) are the same, and similarly for (m / (n + m)). Since we know that n and m are both greater than 0, it follows that n / (n + m) is smaller than 1, and (n / (n + m)) * (n / (n + m)) is smaller than n / (n + m).
Therefore, we can conclude that P(Same color with replacement) (drawing with replacement) is always larger than P(Same color without replacement) (drawing without replacement).
In summary, the probability of drawing balls of the same color without replacement is given by: P(Same color without replacement) = P(Both balls are white) + P(Both balls are black), while the probability of drawing balls of the same color with replacement is: P(Same color with replacement) = (n / (n + m)) * (n / (n + m)) + (m / (n + m)) * (m / (n + m)). And P(Same color with replacement) is always larger than P(Same color without replacement).