# 3.2 Independent and mutually exclusive events  (Page 2/23)

 Page 6 / 23

## Try it

A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing.

1. Compute P ( T ).
2. Compute P ( T | F ).
3. Are T and F independent?.
4. Are F and S mutually exclusive?
5. Are F and S independent?
1. P ( T ) = $\frac{1}{4}$
2. P ( T | F ) = $\frac{1}{2}$
3. No
4. No
5. Yes

Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx (accessed May 2, 2013).

Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013).

## Chapter review

Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent.

In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other.

## Formula review

If A and B are independent, P ( A AND B ) = P ( A ) P ( B ), P ( A | B ) = P ( A ) and P ( B | A ) = P ( B ).

If A and B are mutually exclusive, P ( A OR B ) = P ( A ) + P ( B ) and P ( A AND B ) = 0.

E and F are mutually exclusive events. P ( E ) = 0.4; P ( F ) = 0.5. Find P ( E F ).

J and K are independent events. P ( J | K ) = 0.3. Find P ( J ).

P ( J ) = 0.3

U and V are mutually exclusive events. P ( U ) = 0.26; P ( V ) = 0.37. Find:

1. P ( U AND V ) =
2. P ( U | V ) =
3. P ( U OR V ) =

Q and R are independent events. P ( Q ) = 0.4 and P ( Q  AND  R ) = 0.1. Find P ( R ).

P ( Q AND R ) = P ( Q ) P ( R )

0.1 = (0.4) P ( R )

P ( R ) = 0.25

## Bringing it together

A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News . The factual data are compiled into [link] .

Shirt# ≤ 210 211–250 251–290 290≤
1–33 21 5 0 0
34–66 6 18 7 4
66–99 6 12 22 5

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about P (Shirt# 1–33|≤ 210 pounds)?

The probability that a male develops some form of cancer in his lifetime is 0.4567. The probability that a male has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let C = a man develops cancer in his lifetime and P = man has at least one false positive.

1. P ( C ) = ______
2. P ( P | C ) = ______
3. P ( P | C' ) = ______
4. If a test comes up positive, based upon numerical values, can you assume that man has cancer? Justify numerically and explain why or why not.

1. P ( C ) = 0.4567
2. not enough information
3. not enough information
4. No, because over half (0.51) of men have at least one false positive text

Given events G and H : P ( G ) = 0.43; P ( H ) = 0.26; P ( H AND G ) = 0.14

1. Find P ( H OR G ).
2. Find the probability of the complement of event ( H AND G ).
3. Find the probability of the complement of event ( H OR G ).

Given events J and K : P ( J ) = 0.18; P ( K ) = 0.37; P ( J OR K ) = 0.45

1. Find P ( J AND K ).
2. Find the probability of the complement of event ( J AND K ).
3. Find the probability of the complement of event ( J AND K ).
1. P ( J OR K ) = P ( J ) + P ( K ) − P ( J AND K ); 0.45 = 0.18 + 0.37 - P ( J AND K ); solve to find P ( J AND K ) = 0.10
2. P (NOT ( J AND K )) = 1 - P ( J AND K ) = 1 - 0.10 = 0.90
3. P (NOT ( J OR K )) = 1 - P ( J OR K ) = 1 - 0.45 = 0.55

7.The following data give thenumber of car thefts that occurred in a city in the past 12 days. 63711438726915 Calculate therange, variance, and standard deviation.
express the confidence interval 81.4% ~8.5% in interval form
a bad contain 3 red and 5 black balls another 4 red and 7 black balls, A ball is drawn from a bag selected at random, Find the probability that A is red?
The information is given as, 30% of customers shopping at SHOPNO will switch to DAILY SHOPPING every month on the other hand 40% of customers shopping at DAILY SHOPPING will switch to other every month. What is the probability that customers will switch from A to B for next two months?
Calculate correlation coefficient, where SP(xy) = 144; SS(x) = 739; SS(y) = 58. (2 Points)
The information are given from a randomly selected sample of age of COVID-19 patients who have already survived. These information are collected from 200 persons. The summarized information are as, n= 20; ∑x = 490; s^2 = 40. Calculate 95% confident interval of mean age.
Ashfat
The mode of the density of power of signal is 3.5. Find the probability that the density of a random signal will be more than 2.5.
Ashfat
The average time needed to repair a mobile phone set is 2 hours. If a customer is in queue for half an hour, what is the probability that his set will be repaired within 1.6 hours?
Ashfat
A quality control specialist took a random sample of n = 10 pieces of gum and measured their thickness and found the mean 9 and variance 0.04. Do you think that the mean thickness of the spearmint gum it produces is 8.4
3. The following are the number of mails received in different days by different organizations: Days (x) : 23, 35, 38, 50, 34, 60, 41, 32, 53, 67. Number of mails (y) : 18, 40, 52, 45, 32, 55, 50, 48, 26, 25. i) Fit a regression line of y on x and test the significance of regression. ii) Estimate y
The number of problem creating computers of two laboratories are as follows: Number of computers: 48, 6, 10, 12, 30, 11, 49, 17, 10, 14, 38, 25, 15, 19, 40, 12. Number of computers: 12, 10, 26, 11, 42, 11, 13, 12, 18, 5, 14, 38. Are the two laboratories similar in respect of problem creating compute
Is the severity of the drug problem in high school the same for boys and girls? 85 boys and 70 girls were questioned and 34 of the boys and 14 of the girls admitted to having tried some sort of drug. What can be concluded at the 0.05 level?
null rejected
Pratik
a quality control specialist took a random sample of n=10 pieces of gum and measured their thickness and found the mean 7.6 and standered deviation 0.10. Do you think that the mean thickness of the spearmint gum it produces is 7.5?
99. A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct? a
A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct?
Niaz
what is null Hypothesis
Niaz
what is null Hypothesis
Niaz
when median is greater than mode?
hello
Amaano
is this app useful
Worthy
little bit 😭
G-
oh
Worthy
when tail is positive
Jungjoon
define hypothesis
Worthy
I'm struggling to type it's on my laptop...statistics
Yoliswa
types of averages .mean median mode quarantiles MCQ question
what a consider data?
Out of 25 students, 15 are male. Is the overall proportion of male students 0.7 in AIUB? (4 Points)
15/25=0.6 or 60% standard calculation
Andrea
A quality control specialist took a random sample of n = 10 pieces of gum and measured their thickness and found the mean 7.6 and variance 0.01. Do you think that the mean thickness of the spearmint gum it produces is 7.5? (4 Points)
Omer
10 gums mean = 7.6 variance= 0.01 standard deviation= ? what us the data set?
Andrea
0.6
Rubina By By By Jonathan Long By OpenStax By OpenStax By Rhodes By John Gabrieli By By OpenStax By OpenStax By Bonnie Hurst By OpenStax