A box contains three black balls and two gold balls. A ball is selected at random, its color is recorded, and it is then replaced. A second ball is then selected at random and its color recorded. The outcome associated with this type of selection is an ordered pair (first draw, second draw).
a. List a sample space for this experiment.
b. What is the probability that both balls are black?
c. What is the probability that both balls are gold?
d. Why doesn’t the sum of the probabilities in parts (b) and (c) total 1?
I'm not sure what you mean by sample space.
However, the probability of a gold ball is 2/5 and the probability of a black ball is 3/5.
The probability of two or more events both occurring is found by multiplying the individual probabilities.
They don't total 1, because they don't include the probability of drawing black-gold or gold-black combinations.
I hope this helps.
a. To list the sample space for this experiment, we need to consider all possible outcomes of the first and second draws. The possible outcomes are: (B, B), (B, G), (G, B), (G, G), where B represents a black ball and G represents a gold ball.
b. To calculate the probability that both balls are black, we need to determine the number of favorable outcomes (i.e., both balls are black) and divide it by the total number of possible outcomes. From the sample space in part a, we see that there is only one favorable outcome, which is (B, B). Since there are four possible outcomes in total, the probability of both balls being black is 1/4.
c. Similarly, to calculate the probability that both balls are gold, we again determine the number of favorable outcomes and divide it by the total number of possible outcomes. From the sample space in part a, we see that there is only one favorable outcome, which is (G, G). Therefore, the probability of both balls being gold is also 1/4.
d. The sum of the probabilities in parts (b) and (c) does not total 1 because there is a possibility of the outcomes being different (i.e., one ball being black and the other being gold). In this case, either (B, G) or (G, B) can occur. Since the sample space consists of 4 possible outcomes, and both (B, G) and (G, B) are different outcomes, the sum of the probabilities of parts (b) and (c) would only account for the cases where both balls are of the same color (either black or gold).