math 104 quiz 3 practice problems
1. suppose that you roll three dice. there are 6³ = 216 total outcomes.
(a) what is the probability that the number of dots on the faces of the dice are all the same?
(b) what is the probability that the number of dots on the faces of the dice are not all the same?
(c) what is the probability that the sum of the dots on the faces of the dice is less than or equal to 5?
2. you flip a coin 5 times what is the probability that you get at least 2 heads? for this example there are 2⁵ = 32 outcomes. you should find the probability of the complement of this event and then subtract its probability from 1. this is the formula p(e) + p(e)=1.
3. you roll a six - sided die twice. what is the probability that the sum of the two rolls is greater than 8?
4. suppose that tomorrows weather forecast calls for rain with probability .55, temperatures above 40 with probability .65, and rain with temperatures 40 degrees or below with probability .25.
(a) what is the probability that it will not rain and the temperature will be above 40?
(b) what is the probability that is will not rain and the temperature will be 40 degrees or below.?
5. suppose that e and f are events in a sample space s with p(f)=.34, p(e∩f)=.22, p(e∪f)=.75 determine the following.
(a) p(e)
(b) p(e∩f)
(c) p((e∩f))

Question

Accepted Answer

### 1.
#### (a)
# Explanation:
## Step1: Count favorable outcomes
There are 6 cases where the numbers on the three - dice are the same: (1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6).
## Step2: Calculate probability
The probability formula is $P(A)=\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}}$. Here, the total number of outcomes is $6^3 = 216$. So $P=\frac{6}{216}=\frac{1}{36}$.
# Answer:
$\frac{1}{36}$

#### (b)
# Explanation:
## Step1: Use complementary probability
The event that the numbers on the dice are not all the same is the complement of the event that the numbers on the dice are all the same. Let $A$ be the event that the numbers are all the same and $A'$ be the event that the numbers are not all the same. We know that $P(A')=1 - P(A)$.
## Step2: Substitute value of $P(A)$
Since $P(A)=\frac{1}{36}$, then $P(A')=1-\frac{1}{36}=\frac{35}{36}$.
# Answer:
$\frac{35}{36}$

#### (c)
# Explanation:
## Step1: List favorable combinations
The possible combinations of three - dice values $(x,y,z)$ such that $x + y+z\leq5$ are: $(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)$. The number of favorable outcomes is 10.
## Step2: Calculate probability
Using the formula $P=\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}}$, with total outcomes $n = 216$, we get $P=\frac{10}{216}=\frac{5}{108}$.
# Answer:
$\frac{5}{108}$

### 2.
# Explanation:
## Step1: Find probability of complement event
The complement of getting at least 2 heads in 5 coin - flips is getting 0 or 1 head. The number of ways to get 0 heads (all tails) in 5 flips is $C(5,0)=\frac{5!}{0!(5 - 0)!}=1$. The number of ways to get 1 head in 5 flips is $C(5,1)=\frac{5!}{1!(5 - 1)!}=5$. So the number of ways to get 0 or 1 head is $1 + 5=6$.
## Step2: Calculate probability of complement
The total number of outcomes is $2^5 = 32$. So the probability of the complement event $P(	ext{0 or 1 head})=\frac{6}{32}=\frac{3}{16}$.
## Step3: Calculate probability of original event
Using $P(E)=1 - P(E')$, we have $P(	ext{at least 2 heads})=1-\frac{3}{16}=\frac{13}{16}$.
# Answer:
$\frac{13}{16}$

### 3.
# Explanation:
## Step1: Find total number of outcomes
When rolling a six - sided die twice, the total number of outcomes is $n = 6	imes6=36$.
## Step2: List favorable combinations
The pairs $(x,y)$ such that $x + y>8$ are: $(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)$. The number of favorable outcomes is 10.
## Step3: Calculate probability
Using the formula $P=\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}}$, we get $P=\frac{10}{36}=\frac{5}{18}$.
# Answer:
$\frac{5}{18}$

### 4.
#### (a)
# Explanation:
## Step1: Use set - theoretic approach
Let $R$ be the event of rain and $T$ be the event of temperature above 40. We know that $P(R) = 0.55$, $P(T)=0.65$, $P(R\cap T^c)=0.25$. First, find $P(R\cap T)=P(R)-P(R\cap T^c)=0.55 - 0.25 = 0.3$. Then, using the formula $P(R^c\cap T)=P(T)-P(R\cap T)$, we substitute the values.
## Step2: Calculate probability
$P(R^c\cap T)=0.65 - 0.3=0.35$.
# Answer:
$0.35$

#### (b)
# Explanation:
## Step1: Calculate $P(R^c)$
We know that $P(R) = 0.55$, so $P(R^c)=1 - P(R)=1 - 0.55 = 0.45$. Also, $P(R\cap T^c)=0.25$.
## Step2: Calculate probability
The probability that it will not rain and the temperature will be 40 or below is $P(R^c\cap T^c)=P(R^c)-P(R^c\cap T)$. Since $P(R^c\cap T) = 0.35$ (from part (a)), then $P(R^c\cap T^c)=0.45-0.35 = 0.1$.
# Answer:
$0.1$

### 5.
#### (a)
# Explanation:
## Step1: Use the formula $P(E\cup F)=P(E)+P(F)-P(E\cap F)$
We know that $P(F) = 0.34$, $P(E'\cap F)=0.22$, so $P(E\cap F)=P(F)-P(E'\cap F)=0.34 - 0.22 = 0.12$. Also, $P(E\cup F)=0.75$.
## Step2: Solve for $P(E)$
Substituting into the formula $P(E\cup F)=P(E)+P(F)-P(E\cap F)$, we get $0.75=P(E)+0.34 - 0.12$. Rearranging gives $P(E)=0.75-(0.34 - 0.12)=0.53$.
# Answer:
$0.53$

#### (b)
# Explanation:
We already found in part (a) that $P(E\cap F)=P(F)-P(E'\cap F)$.
# Answer:
$0.12$

#### (c)
# Explanation:
## Step1: Use the formula $P((E\cap F)')=1 - P(E\cap F)$
Since $P(E\cap F)=0.12$, then $P((E\cap F)')=1 - 0.12 = 0.88$.
# Answer:
$0.88$

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QUESTION IMAGE

Question

Explanation:

1.

(a)

Step1: Count favorable outcomes

Step2: Calculate probability

Step1: Use complementary probability

Step2: Substitute value of $P(A)$

Step1: List favorable combinations

Step2: Calculate probability

Answer:

(b)