A box contains three black balls and two golden balls. A ball is selected at random, it's color is recorded and it is then replaced. A second ball is selected at random, it's color is recorded it is then replaced. The outcome is associated with this type of selection is an ordered pair.
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 dosent the sum of the probabilities in part B and C total to 1?
Can anyone help me get started?
Sure! Let's break down the problem step by step.
A. To list a sample space for this experiment, we need to consider all the possible outcomes. Since there are two selections, we can represent the outcomes as ordered pairs. Let's use "B" to represent a black ball and "G" to represent a gold ball. The possible outcomes are:
( B , B )
( B , G )
( G , B )
( G , G )
B. To find the probability that both balls are black, we need to identify the number of favorable outcomes (both balls are black) and divide it by the total number of possible outcomes. In the sample space we listed above, there is only one outcome where both balls are black: ( B , B ). Therefore, the probability is 1/4 or 0.25.
C. Similarly, to find the probability that both balls are gold, we need to identify the number of favorable outcomes (both balls are gold) and divide it by the total number of possible outcomes. In the sample space we listed, there is also only one outcome where both balls are gold: ( G , G ). Therefore, the probability is 1/4 or 0.25.
D. The sum of the probabilities in parts B and C does not total to 1 because there are other possible outcomes besides both balls being black or both balls being gold. The sample space also includes outcomes where the first ball is black and the second ball is gold (B, G) or the first ball is gold and the second ball is black (G, B). These outcomes are not included in the probabilities calculated in parts B and C.
I hope this explanation helps you understand how to solve the problem.