In a class of students, the following data table summarizes how many students have a cat or a dog. What is the probability that a student who does not have a dog has a cat?
Has a cat Does not have a cat
Has a dog 9 12
Does not have a dog 3 5
The probability that a student who does not have a dog has a cat can be found by taking the number of students who have a cat but do not have a dog (which is 12) and dividing by the total number of students who do not have a dog (which is 3+5 = 8). Therefore, the probability is:
12/8 = 3/2 or 1.5
However, probabilities cannot be greater than 1, so the answer is not valid. This might be because of some mistake in the data or in the calculation. It could be that there is some overlap between the two categories (some students could have both a cat and a dog), or that the data is incomplete. In any case, without more information, we cannot give a valid probability for this question.
The following table represents the highest educational attainment of all adult residents in a certain town. If an adult is chosen randomly from the town, what is the probability that they have a high school degree or some college, but have no college degree? Round your answer to the nearest thousandth.
Age 20-29 Age 30-39 Age 40-49 Age 50 & over Total
High school only 931 541 855 1242 3569
Some college 2306 1873 1474 2392 8045
Bachelor's degree 719 437 1189 1116 3461
Master's degree 885 656 308 1324 3173
Total 4841 3507 3826 6074 18248
The number of adults who have a high school degree or some college but no college degree is the sum of the values in the "High school only" and "Some college" rows, but not in the "Bachelor's degree" or "Master's degree" rows. Therefore:
High school degree or some college, no college degree = 931 + 541 + 855 + 2306 + 1873 + 1474
= 7979
The total number of adults in the town is 18,248, so the probability that an adult chosen randomly from the town has a high school degree or some college, but no college degree is:
7979 / 18248 ≈ 0.437
Rounding to the nearest thousandth, the answer is 0.437.
The following table represents the highest educational attainment of all adult residents in a certain town. If a resident who is 30-49 years old is chosen at random, what is the probability that they have completed a bachelor's or master's degree? Round your answer to the nearest thousandth.
Age 20-29 Age 30-39 Age 40-49 Age 50 & over Total
High school only 992 816 1155 1266 4229
Some college 1947 912 1275 2171 6305
Bachelor's degree 1197 1320 1272 614 4403
Master's degree 1055 774 213 872 2914
Total 5191 3822 3915 4923 17851
To find the probability that a resident who is 30-49 years old has completed a bachelor's or master's degree, we need to add up the number of residents in the "Bachelor's degree" and "Master's degree" rows that fall in the "Age 30-39" and "Age 40-49" columns:
Bachelor's or master's degree, 30-49 years old = 1272 + 1320 + 213 + 774 = 2579
The total number of residents who are 30-49 years old is the sum of the values in the "Age 30-39" and "Age 40-49" columns:
Total residents, 30-49 years old = 3822 + 3915 = 7737
Therefore, the probability that a resident who is 30-49 years old has completed a bachelor's or master's degree is:
2579 / 7737 ≈ 0.333
Rounding to the nearest thousandth, the answer is 0.333.
A lock has a code of 3 numbers from 1 to 25. If no numbers in the code are allowed to repeat, how many different codes could be made?
There are 25 options for the first number in the code, since it can be any number from 1 to 25. For the second number, there are only 24 options left, since one number has already been used. Similarly, for the third number, there are only 23 options left. Therefore, the total number of possible codes is:
25 × 24 × 23 = 13,800
So there are 13,800 different codes that could be made.
Ms. Bell's mathematics class consists of 6 sophomores, 11 juniors, and 13 seniors. How many different ways can Ms. Bell create a 5-member committee of seniors if each senior has an equal chance of being selected?
There are 13 seniors in the class, and we need to choose a committee of 5. We can use the combination formula to calculate the number of ways to do this:
C(13, 5) = 13! / (5! * (13-5)!) = 1287
Therefore, there are 1,287 different ways for Ms. Bell to create a 5-member committee of seniors.