1. A bag contains 12 red checkers and 6 black checkers. You will randomly select two checkers, one at a time, without replacement.
(a) In how many ways can you select two checkers so that at least one of the checkers is red? List the ways.
(b) What is the probability that you select a black checker and then a red checker? Show your work.
Matt got caught lmao
Matt, don't feel bad. I have a student taking the same class and despite the fact we have watched class connects and participated in lessons, we are still not sure of how to do this either. At least you are RESEARCHING how to do it instead of not turning in the assignment. Because, unless you are going into pre-med, engineering, design, genetics, etc, I can't possibly fathom giving a student a zero for attempting an assignment. Perhaps he isn't trying to cheat, but gain understanding of how to do the problem....which tells me TEACHER, there is a failure ratio somewhere. Especially when the format of education currently presented by GCA falls well below my expectations. We won't be using this again. I spend far too much time using ALEKS, Kahn Academy videos and other sources....rather than GCA's.
(a) To find the number of ways to select two checkers such that at least one of them is red, we can use the principle of complementary counting.
First, let's calculate the total number of ways to choose any two checkers from the bag without regard to their color. This can be done using the combination formula, denoted as "nCr," which represents the number of ways to choose r items from a set of n items without replacement.
In this case, we have 18 checkers in total (12 red + 6 black). Therefore, the total number of ways to select two checkers without regard to their color is given by:
Total ways = 18C2 = (18!)/(2!(18-2)!) = (18!)/(2!16!) = (18 * 17) / (2 * 1) = 153
To determine the number of ways to select two checkers such that at least one of them is red, we need to subtract the number of ways to select two black checkers from the total ways.
Number of ways to select two black checkers = 6C2 = (6!)/(2!(6-2)!) = (6!)/(2!4!) = (6 * 5) / (2 * 1) = 15
Therefore, the number of ways to select two checkers such that at least one of them is red is given by:
Number of ways = Total ways - Number of ways to select two black checkers = 153 - 15 = 138
Hence, there are 138 ways to select two checkers such that at least one of them is red.
To list all the ways, we can perform calculations using combinations:
- One red and one black: 12C1 * 6C1 = (12*6) = 72 ways
- Two reds: 12C2 = 66 ways
Therefore, there are 138 ways as follows:
- One red and one black: 72 ways
- Two reds: 66 ways
(b) To find the probability of selecting a black checker, followed by a red checker, we need to calculate the probability of each event and then multiply them together.
First, let's calculate the probability of selecting a black checker. Since there are 18 checkers in total and 6 of them are black, the probability of selecting a black checker on the first draw is given by:
Probability of drawing a black checker = Number of black checkers / Total number of checkers = 6 / 18 = 1/3
After selecting a black checker, there are 17 checkers remaining, out of which 12 are red. Therefore, the probability of selecting a red checker on the second draw, given that a black checker was selected on the first draw, is given by:
Probability of drawing a red checker = Number of red checkers (after the first draw) / Remaining number of checkers = 12 / 17
To find the overall probability, we multiply the probability of each event together:
Probability of selecting a black checker and then a red checker = Probability of drawing a black checker * Probability of drawing a red checker
= (1/3) * (12/17)
= 12/51
= 4/17
= 0.2353 (rounded to 4 decimal places)
Therefore, the probability of selecting a black checker and then a red checker is 4/17 or approximately 0.2353.