What is a line segment

jimmy is moving to a new apartment. he fills a box with 4 books, 5 DVDs, and 2 CD's. when he starts unpacking his new place he p

pav-90 [236]

Answer:

$\frac{2}{11}$

Step-by-step explanation:

First, find the probability of him pulling out a book.

There are 11 objects in total, and there are 4 books.

So, the probability of pulling out a book is $\frac{4}{11}$

Next, find the probability of him pulling out a DVD after.

Since a book was taken out, there are only 10 objects left. There are 5 DVDs.

So, the probability of pulling out a DVD is $\frac{5}{10}$ , or simplified to $\frac{1}{2}$

To find the probability that this happens in order, multiply the probabilities:

$\frac{4}{11}$ x $\frac{1}{2}$

= $\frac{2}{11}$

So, the probability is $\frac{2}{11}$

7 0

2 years ago

A training company provides classroom and online instruction to hundreds of people per month. As part of their training program

Lynna [10]

Answer:

According to the sample data, we can claim that the current satisfaction rate is different from last year's.

Step-by-step explanation:

The explanation is in the picture.

3 0

3 years ago

What is 2M? <br> Is anyone good at Matrices?

goldenfox [79]

Answer:

It's the first one

Step-by-step explanation:

Because 2M means you are multiplying all the values of M with 2

3 0

2 years ago

48 pens are packed in four boxes how many pens are packed in nine boxes

katovenus [111]

Answer:

108 pens

Step-by-step explanation:

48/4=x/9

4x=432

x=108

3 0

3 years ago

Read 2 more answers

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(c) $\displaystyle P(B) = \frac{9}{100}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

2 years ago