Question

1. What is the sample space if you roll two dice? What is the probability of rolling a sum of 7?
2. You and a friend go to a restaurant where you both order a soup and a salad. There are four soups (A, B, C, D) and 3
salads (1, 2, 3) to choose from. If you and your friend decide to get completely unique meals, what is the sample space
of possible orders you can make?
3. Give 3 examples of disjoint events.
4. The probability that Bob gets to work on time is 80%. The probability that Sarah gets to work on time is 90%. The
probability that they're both on time is 72%. What's the probability that Bob OR Sarah is late for work?
5. Jake's Pizza Place is offering a deal where they'll deliver your pizza in under 30 minutes or your pizza is free. They
manage to deliver the pizza in under 30 minutes 92% of the time. What's the probability that you'll get your pizza for
free?
6. The probability that a randomly selected student in our class likes soccer is 54%. If we select 3 random students
from our class, what is the probability that all of them like soccer?
7. The probability that a randomly selected student in our class likes soccer is still 54%. If we select 4 random students
from our class, what is the probability that all of them don't like soccer?
8. There are 6 blue marbles, 5 red marbles, and 4 green marbles in a box. If you pull a green marble out of the box, what
is the probability that the next marble you draw is red?
9.60% of students in a class studied for a test AND got an A on that test. 75% of students studied for the test. What is
the probability that a student got an A given that they studied for the test?

Answer 1

Probabilities are used to determine how likely it is for an event to happen.

(1) A roll of two dice

Let the dice be A and B. The sample space of each is:

$A = \{1,2,3,4,5,6\}\\B = \{1,2,3,4,5,6\}$

To get the sample space (S) of the sum of the dice, we simply add up the individual element of both dice. So, we have:

$S = \{(1+1),(1+2),(1+3),(1+4),(1+5),(1+6),.............,(6+1),(6+2),(6+3),(6+4),(6+5),(6+6)\}$

This gives

$S = \{2,3,4,5,6,7,.............,7,8,9,10,11,12\}$

The complete sample space is:

$S = \{2,3,3,4,4,4,5,5,5,5,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,8,9,9,9,9,10,10,10,11,11,12\}$

$n(S) = 36$ --- the total outcomes

The probability of rolling a sum of 7 is:

$P(7) = \frac{n(7)}{n(S)}$

Where

$n(7) = 6$ ---- number of 7's in the sample space

So, we have:

$P(7) = \frac 6{36}$

$P(7) = 0.167$

The probability of rolling a sum of 7 is 0.167

(2) Restaurant

We have:

$Soup = \{A,B,C,D\}$

$Salads = \{1,2,3\}$

To get the sample space (S), we simply pick the individual elements of the above data. So, we have:

$S = \{A1, A2, A3,B1, B2, B3,C1, C2, C3,D1, D2, D3\}$

(3) Disjoint events

Events are said to be disjointed if they cannot happen at the same time. 3 examples are:

• Event 1: You are in China. Event 2: You are in Nigeria
• Event 1: Heading towards north. Event 2: Heading towards south
• Even 1: Outcome of even number in a roll of die. Event 2: Outcome of an odd number in the same die

(4) Bob and Sarah

Let

$A \to$ Bob gets to work on time

$B \to$ Sarah gets to work on time

$P(A) = 80\% \\ P(B) = 90\%\\P(A\ n\ B) = 72\%$

The probability that Bob OR Sarah is late for work is represented as:

$P(A'\ or\ B')$

This is calculated using:

$P(A'\ or\ B') = P(A') + P(B') - P(A'\ n\ B')$

Where:

$P(A') = 1 - P(A) = 1 - 80\% = 20\%$

$P(B') = 1 -P(B) = 1 - 90\% =10\%$

$P(A\ n\ B') = P(A) \times P(B') = 20\% \times 10\% = 2\%$

So, we have:

$P(A'\ or\ B') = P(A') + P(B') - P(A'\ n\ B')$

$P(A'\ or\ B') = 20\% + 10\% - 2\%$

$P(A'\ or\ B') = 28\%$

$P(A'\ or\ B') = 0.28$

The probability that Bob OR Sarah is late for work is 0.28

(5) Jake's Pizza

Let

$A \to$ The proportion of time they deliver on time

$B \to$ You get your pizza for free

So:

$P(A) = 92\%$

You get your pizza for free if they didn't deliver within 30 minutes.

This means that:

$P(B) = 1 - P(A)$ --- Complement rule

So, we have:

$P(B) = 1 - 92\%$

$P(B) = 0.08$

The probability that you get your pizza for free is 0.08.

(6) Students and Soccer

Let:

$A \to$ Probability that a student likes soccer

So:

$P(A) = 54\%$

If all 3 selected students like soccer, the probability is calculated as:

$Pr = P(A) \times P(A) \times P(A)$

$Pr = P(A)^3$

$Pr = 54\%^3$

$Pr = 0.157$

The probability that all 3 selected students like soccer is 0.157

(7) Students and Soccer

Let:

$A \to$ Probability that a student likes soccer

$B \to$ Probability that a student does not like soccer

So:

$P(A) = 54\%$

$P(B)= 1 - P(A)= 1 - 54\% = 46\%$ ---- Complement rule

If all 4 selected students do not like soccer, the probability is calculated as:

$Pr = P(B) \times P(B) \times P(B) \times P(B)$

$Pr = P(B)^4$

$Pr = 46\%^4$

$Pr = 0.045$

The probability that all 4 selected students do not like soccer is 0.045

(8) Marbles

$Blue = 6 \\ Green = 4 \\ Red = 5$

When the green marble is selected, we are left with

$Blue = 6 \\ Green = 3 \\ Red = 5$

The probability of selecting a red marble at this point is:

$P(Red) = \frac{Red}{Blue + Green + Red}$

$P(Red) = \frac{5}{6+3+5}$

$P(Red) = \frac{5}{14}$

$P(Red) = 0.357$

The probability that the next marble you draw is red is 0.357

(9) Studying for test

Let

$A \to$ Students got an A

$B \to$ Students studied for test

So:

$P(A\ n\ B) = 60\% \\ P(B) = 75\%$

The probability that a student got an A given that they studied for the test is  represented as:

$P(A | B)$

So, we have:

$P(A | B) = \frac{P(A\ n\ B)}{P(B)}$

$P(A | B) = \frac{60\%}{75\%}$

$P(A | B) = 0.80$

The probability that a student got an A given that they studied for the test is  0.80

Read more about probabilities at:

https://brainly.com/question/16693319