math 104 quiz 3 practice problems\n1. suppose that you roll three dice. there are 6³ = 216 total…

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 $P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. 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 $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$, 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 $P(\text{0 or 1 head})=\frac{6}{32}=\frac{3}{16}$.

Step3: Calculate probability of original event

Using $P(E)=1 - P(E')$, we have $P(\text{at least 2 heads})=1-\frac{3}{16}=\frac{13}{16}$.

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 $n = 6\times6=36$.

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 $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$, 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:

$0.88$