Answer:
a) S = {1, 2, 3}
b) P(odd number) = 
c) No
d) Yes
Step-by-step explanation:
a) The sample space is the set of all possible outcomes. By definition, the elements of a set should not be repeated. Hence, the sample space S = {1, 2, 3}
However, the sample is not equiprobable because each element has different probabilities.
b) P(odd number) = 
Note that the odd numbers are 1 (on three faces) and 3 (on one face).
c) The fact the die has been biased does not change the possible outcomes. It only changes the probability of getting any given number.
d) Because the 3-face has been loaded, this probability changes. In fact, it is calculated thus:
Let's assume the probability for 1 or 2 is 
. Then that of 3 is 
(because it is twice the others). The sum of probabilities must be 1.
P(odd number) = 
Prob(1) + Prob(3)
= 
= 
Answer: 16) 
(17) 
(18) 
(19) 3x - 18
<u>Step-by-step explanation:</u>
Inverse is when you swap the x's and y's and then solve for y
16) Given: y = x²
Swapped: x = y²
17) Given: y = 2x + 4
Swapped: x = 2y + 4
x - 4 = 2y
18) Given: y = (x + 2)²
Swapped: x = (y + 2)²
√x = y + 2
19) Given: 
Swapped: 
3(x - 6) = y
3x - 18 = y
1. 1/4, 2/7, 3/8
2. 12/5
3. 3/4, 7/10, 7/12
4. 1/6, 1/2, 1/4
Answer:
34.87% probability that all 5 have a wireless device
Step-by-step explanation:
For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
81% of students own a wireless device.
This means that 
If 5 students are selected at random, what is the probability that all 5 have a wireless device?
This is P(X = 5) when n = 5. So
34.87% probability that all 5 have a wireless device
Answer:
This answer of this question is Yes
