Exploring Probability Unit Test Part 2
Note: In the space provided, enter your response and outline all the methods you used to resolve this issue.
A number cube is rolled twice. Then, find P(not 2 if even). In its simplest form, express the probability as a fraction.
Solution:
First, we need to determine the sample space of rolling a number cube twice. Since each roll is independent, we can use the multiplication rule of counting to find the total number of outcomes:
Sample space = {1,1}, {1,2}, {1,3}, {1,4}, {1,5}, {1,6},
{2,1}, {2,2}, {2,3}, {2,4}, {2,5}, {2,6},
{3,1}, {3,2}, {3,3}, {3,4}, {3,5}, {3,6},
{4,1}, {4,2}, {4,3}, {4,4}, {4,5}, {4,6},
{5,1}, {5,2}, {5,3}, {5,4}, {5,5}, {5,6},
{6,1}, {6,2}, {6,3}, {6,4}, {6,5}, {6,6}
There are 36 outcomes in the sample space.
Next, we need to find the events "even" and "not 2".
The event "even" is the set of outcomes that have an even number on at least one roll. This can be found by counting the number of outcomes that have 2, 4, or 6 on at least one roll:
{2,1}, {2,2}, {2,3}, {2,4}, {2,5}, {2,6},
{4,1}, {4,2}, {4,3}, {4,4}, {4,5}, {4,6},
{6,1}, {6,2}, {6,3}, {6,4}, {6,5}, {6,6}
There are 18 outcomes in the event "even".
The event "not 2" is the set of outcomes that do not have a 2 on either roll. This can be found by counting the number of outcomes that do not have 2 on either roll:
{1, 1}, {1, 3}, {1, 4}, {1, 5}, {1, 6},
{3, 1}, {3, 3}, {3, 4}, {3, 5}, {3, 6},
{4, 1}, {4, 3}, {4, 4}, {4, 5}, {4, 6},
{5, 1}, {5, 3}, {5, 4}, {5, 5}, {5, 6},
{6, 1}, {6, 3}, {6, 4}, {6, 5}, {6, 6}
There are 25 outcomes in the event "not 2".
To find the probability of "not 2 if even", we need to find the likelihood of an outcome being in both events "even" and "not 2". This can be found by finding the intersection of the two events:
{4, 1}, {4, 3}, {4, 4}, {4, 5}, {4, 6},
{6, 1}, {6, 3}, {6, 4}, {6, 5}, {6, 6}
There are 10 outcomes in the intersection of "even" and "not 2".
Therefore, the probability of "not 2 if even" is:
P(not 2 if even) = number of outcomes in the intersection of "even" and "not 2" / number of outcomes in "even"
= 10 / 18
In its simplest form, the probability is 5/9.
Marissa is researching information about martial arts students. She found that 7 out of 12 martial artists practice every day. There are 144 martial arts students at a school.
a. Predict how many students practice every day.
b. What is the sample size?
a. To predict how many students practice every day, we can use a proportion. We know that 7 out of 12 martial artists practice every day, so we can write:
7/12 = x/144
where x is the number of martial arts students who practice every day.
To solve for x, we can cross-multiply and simplify:
12x = 7 * 144
x = 84
Therefore, we predict that 84 martial arts students practice every day.
b. The sample size is the total number of martial arts students at the school, which is given as 144.
You roll a number cube twice. Find P(even, then not 2). Write the probability as a fraction in simplest form.
There are six possible outcomes when rolling a number cube: 1, 2, 3, 4, 5, or 6.
For the first roll, there are three even numbers: 2, 4, and 6. Therefore, the probability of rolling an even number on the first roll is 3/6, or 1/2.
For the second roll, there are five numbers left after rolling the first number (since we can't roll the same number twice), but only one of them is not 2: 1, 3, 4, 5, and 6. Therefore, the probability of rolling a number that is not 2 on the second roll is 1/5.
Since the rolls are independent, we can multiply the probabilities to find the probability of rolling an even number on the first roll AND rolling a number that is not 2 on the second roll:
P(even, then not 2) = P(even) x P(not 2)
P(even, then not 2) = 1/2 x 1/5
P(even, then not 2) = 1/10
Therefore, the probability of rolling an even number on the first roll and a number that is not 2 on the second roll is 1/10.
A number cube is rolled 450 times. The number 3 comes up 67 times.
a. What is the theoretical probability of rolling a 3? Write your answer as a fraction in simplest form.
b. What is the experimental probability of rolling a 3? Write your answer as a fraction in simplest form.
a. The theoretical probability of rolling a 3 on a number cube is 1/6, since there is one favorable outcome (rolling a 3) out of six equally likely possible outcomes. Therefore, the theoretical probability of rolling a 3 is:
P(rolling a 3) = 1/6
b. The experimental probability of rolling a 3 is found by dividing the number of times a 3 was rolled by the total number of rolls:
Experimental probability = number of times rolling a 3 / total number of rolls
Experimental probability = 67 / 450
To simplify this fraction, we can divide both the numerator and denominator by 5:
Experimental probability = 13 / 90
Therefore, the experimental probability of rolling a 3 is 13/90.
You mix the letters M, A, T, H, E, M, A, T, I, C, A, and L thoroughly. Without looking, you draw one letter. Find the probability P(A). Write the probability as:
a. a fraction in simplest form
b. a decimal
c. a percent
There are a total of 12 letters, with 2 of them being A.
a. The probability of drawing an A can be written as:
P(A) = number of A's / total number of letters
P(A) = 2 / 12
To simplify this fraction, we can divide both the numerator and denominator by 2:
P(A) = 1 / 6
Therefore, the probability of drawing an A is 1/6.
b. To write this probability as a decimal, we can divide the numerator by the denominator:
P(A) = 2 / 12 = 0.166666...
Rounded to the nearest hundredth, this is 0.17.
c. To write this probability as a percentage, we can multiply the decimal by 100:
P(A) = 0.166666... * 100 = 16.666...
Rounded to the nearest tenth, this is 16.7%.
Therefore, the probability of drawing an A is 1/6, 0.17, or 16.7%.
To find the probability of an event, we need to determine the total number of possible outcomes and the number of favorable outcomes.
In this scenario, a number cube (a regular six-sided die) is rolled twice. Each roll has six possible outcomes, ranging from 1 to 6.
First, let's determine the total number of possible outcomes when rolling the number cube twice. Since each roll has six possible outcomes, the total number of outcomes will be 6 multiplied by 6, which equals 36.
Next, we need to find the number of favorable outcomes, specifically the event of not getting a 2 if the number rolled is even.
If the number rolled is even, the possible outcomes are 2, 4, and 6. We want to find the probability of not getting a 2 from these three possibilities. Hence, we subtract 1 (the outcome of not getting a 2) from the total number of even outcomes.
So, the number of favorable outcomes is 3 - 1 = 2.
Now, we can find the probability by dividing the number of favorable outcomes by the total number of possible outcomes.
P(not 2 if even) = Number of favorable outcomes / Total number of possible outcomes
P(not 2 if even) = 2 / 36
To express the probability as a simplified fraction, we need to divide both the numerator and denominator by their greatest common factor, which is 2.
P(not 2 if even) = 1 / 18
Therefore, the probability of not rolling a 2 if the number rolled is even is 1/18.