1.M&M candies are great for probability. The following tables are the color distributions for the candies. Fill in each table with the missing probability and answer the questions that follow.
Plain Brown Blue Green Orange Red Yellow
Probability 0.3 0.1 0.1 0.1 0.02 ???
Peanut Brown Blue Green Orange Red Yellow
Probability 0.2 0.3 0.1 0.1 0.01 ???
a) What is the probability that a plain M&M is red or blue? = 0.02+0.1=0.12
b) b) What is the probability that a peanut M&M is not brown? = 0.01+0.3=0.31
2. In order to ensure the safety of school classrooms the local Fire Marshall does an inspection at Thomas Jefferson High School every month, looking for faulty wiring, overloaded circuits, etc. At TJHS the new Academic Wing has 5 math rooms, 10 science rooms, and 10 English rooms. The science rooms are divided into 8 biology and 2 chemistry rooms. Each month, the Fire Marshall randomly picks one of the rooms in the new wing to inspect each month. Define the following events:
S = the event the selected room is a science room
B = the event the selected room is a biology room
M = the event the selected room is a math room
E = the event the selected room is an English room
C = the event the selected room is a chemistry room
Calculate the probabilities of the events described below:
a) P(S) - 10/25
b) P(M or E) - 15/25
c) P(E or B -18/25
) d) P(S and not C) -8/25
3. Research shows that the probability of dying from heart disease is o.45 and the probability of dying from cancer is 0.22.
a. What is the probability that a death was due to either heart disease or cancer 0.45+.22= 67 or 67%
b. What is the probability that a death was due to some other cause? 100-67=33 or 33%
To fill in the missing probabilities in the tables, we need to ensure that the sum of the probabilities for each table is equal to 1.
In the first table for plain M&M candies, the probabilities given are: Plain Brown - 0.3, Blue - 0.1, Green - 0.1, Orange - 0.1, Red - 0.02. To find the missing probability for Yellow, we subtract the sum of the other probabilities from 1: 1 - (0.3 + 0.1 + 0.1 + 0.1 + 0.02) = 0.38. Therefore, the missing probability for Yellow is 0.38.
Similarly, in the second table for peanut M&M candies, the probabilities given are: Peanut Brown - 0.2, Blue - 0.3, Green - 0.1, Orange - 0.1, Red - 0.01. To find the missing probability for Yellow, we subtract the sum of the other probabilities from 1: 1 - (0.2 + 0.3 + 0.1 + 0.1 + 0.01) = 0.29. Therefore, the missing probability for Yellow is 0.29.
Now, let's answer the questions using the completed tables:
a) What is the probability that a plain M&M is red or blue?
To find the probability, we add the probabilities of red and blue: 0.02 + 0.1 = 0.12. Therefore, the probability that a plain M&M is red or blue is 0.12.
b) What is the probability that a peanut M&M is not brown?
To find the probability, we add the probabilities of all the colors except brown: 0.01 + 0.3 + 0.1 + 0.1 + 0.29 = 0.8. Therefore, the probability that a peanut M&M is not brown is 0.8.
Moving on to the second question:
For Thomas Jefferson High School (TJHS), we have the following information:
- Math rooms: 5
- Science rooms: 10 (8 biology, 2 chemistry)
- English rooms: 10
Define the following events:
S = the event the selected room is a science room
B = the event the selected room is a biology room
M = the event the selected room is a math room
E = the event the selected room is an English room
C = the event the selected room is a chemistry room
a) P(S) - The probability of selecting a science room is given by the total number of science rooms divided by the total number of rooms in the new wing:
P(S) = 10 / (5 + 10 + 10) = 10 / 25 = 0.4. Therefore, the probability that the selected room is a science room is 0.4.
b) P(M or E) - The probability of selecting a math room OR an English room is given by the sum of the number of math rooms and the number of English rooms divided by the total number of rooms in the new wing:
P(M or E) = (5 + 10) / (5 + 10 + 10) = 15 / 25 = 0.6. Therefore, the probability that the selected room is a math room OR an English room is 0.6.
c) P(E or B) - The probability of selecting an English room OR a biology room is given by the sum of the number of English rooms and the number of biology rooms divided by the total number of rooms in the new wing:
P(E or B) = (10 + 8) / (5 + 10 + 10) = 18 / 25 = 0.72. Therefore, the probability that the selected room is an English room OR a biology room is 0.72.
d) P(S and not C) - The probability of selecting a science room AND not selecting a chemistry room is given by the number of science rooms (excluding the chemistry rooms) divided by the total number of rooms in the new wing:
P(S and not C) = 8 / (5 + 10 + 10) = 8 / 25 = 0.32. Therefore, the probability that the selected room is a science room AND not a chemistry room is 0.32.
Moving on to the third question:
a) What is the probability that a death was due to either heart disease or cancer?
To find the probability, we add the probabilities of heart disease and cancer: 0.45 + 0.22 = 0.67, or 67%. Therefore, the probability that a death was due to either heart disease or cancer is 67%.
b) What is the probability that a death was due to some other cause?
To find the probability, we subtract the probability of deaths due to heart disease and cancer from 1: 1 - 0.67 = 0.33, or 33%. Therefore, the probability that a death was due to some other cause is 33%.