Probability

Revision 4

Question 1

An urn has two white balls and two black balls in it. Two balls are drawn out without replacing the first ball.a. What is the probability that the second ball is white, given that the first ball is white?b. What is the probability that the first ball was white, given that the second ball is white?

Question 1

An urn has two white balls and two black balls in it. Two balls are drawn out without replacing the first ball.a. What is the probability that the second ball is

white, given that the first ball is white?b. If the first ball is white, there are now 3 balls

left and only one is white: 1/3

Question 1

An urn has two white balls and two black balls in it. Two balls are drawn out without replacing the first ball.b. What is the probability that the first ball was white, given that the second ball is white?

Question 2

Draw Venn diagrams

P(A)=0.4, P(B)=0.6, P(A B)=0.3, Find

0.1 0.3 0.3

0.3

P(A)=0.3, P(B)=0.4,Find

0.1 0.2 0.2

0.5

2c. Mutually Exclusive

0.3 0.6

P(A)=0.5,

find the probabilities of the events

P(A)=0.5,

Draw a Venn diagram

0.20.3 0.3

d. P(A)=0.5,

Draw a Venn diagram

0.20.3 0.3

A B

e. P(A)=0.4,

Draw a Venn diagram

0.28

0.42

0.18

A B

0.12

Question 3

What does it mean for two events A and B to be statistically independent?

Question 3

What does it mean for two events A and B to be statistically independent? It means that the occurrence of one does not affect the probability of the other.

Question 4

On a certain type of aircraft the warning lights (showing green for normal and red for trouble) for the engines are accurate 90% of the time. If there are problems with the engines on 2% of all flights, find the probability that there is a fault with an engine, given that the warning light shows red.

T

T’

0.02

0.98

R

RG

G

0.9

0.9

0.1

0.1

We can also create a table: Assume we look at 1000 flights

Red GreenTrouble 20OK 980Totals 1000

“problems with the engines on 2% of flights”

We can also create a table: Assume we look at 1000 flights

Red GreenTrouble 18 20OK 882 980Totals 1000

“warning lights are accurate 90% of the time”

We can also create a table: Assume we look at 1000 flights

Red Green TotalsTrouble 18 2 20OK 98 882 980Totals 116 884 1000

Finish the table

find the probability that there is a fault with an engine, given that the warning light shows red.

Red Green TotalsTrouble 18 2 20OK 98 882 980Totals 116 884 1000

Question 5

One of the biggest problems with conducting a mail survey is the poor response rate. In an effort to reduce nonresponse, several different techniques for formatting questionnaires have been proposed. An experiment was conducted to study the effect of the questionnaire layout and page size on response in a mail survey. The results are given below.

a. What proportion of the sample responded to the questionnaire?

a. What proportion of the sample responded to the questionnaire?

b. What proportion of the sample received the typeset small-page version?

c. What proportion of those who received a typeset large-page version actually responded to the questionnaire?

d. What proportion of the sample received a typeset large-page questionnaire and responded?

e. What proportion of those who responded to the questionnaire actually received a type-written large page

questionnaire?

f. By looking at the response rates for each of the four formats, what do you conclude from the study?

f. By looking at the response rates for each of the four formats, what do you conclude from the study?

Type set (Large page) gave the best response rate at 68% with typewritten (large page) almost the same at 66% and Type set (Small page) was the worst at 51%. As a margin of error is likely I would conclude that the response rates seem to be better for large page.

Question 6

A car park contains five Japanese cars and six non Japanese cars. A random variable X is defined by the number of Japanese cars among the first three cars to leave. a. Find the probability distribution of X. b. Calculate the expected number of Japanese cars to leave (among the first three cars to leave).c. Calculate the standard deviation.

Question 6

A car park contains five Japanese cars and six non Japanese cars. A random variable X is defined by the number of Japanese cars among the first three cars to leave. a. Find the probability distribution of X.

Distribution TableX 0 1 2 3

P(X=x) 120/990 450/990 360/990 60/990

Question 6

A car park contains five Japanese cars and six non Japanese cars. A random variable X is defined by the number of Japanese cars among the first three cars to leave. b. Calculate the expected number of Japanese

cars to leave (among the first three cars to leave).

Distribution TableX 0 1 2 3

P(X=x) 120/990=4/33

450/990=5/11

360/990=4/11

60/9902/33

Expected numberX 0 1 2 3

P(X=x) 4/33 5/11 4/11 2/33

VarianceX 0 1 2 3

P(X=x) 4/33 5/11 4/11 2/33

Question 7

The mean salary at Carter-Drumfield sportswear store is $24 000 per annum with a standard deviation of $600. All employees get a $500 rise per annum.a. What will be the new mean salary and standard deviation of salaries?b. If instead of a flat rise, each employee had an income increase of 2%, what would be the new mean salary and standard deviation?

Question 7

The mean salary at Carter-Drumfield sportswear store is $24 000 per annum with a standard deviation of $600. All employees get a $500 rise per annum.a. What will be the new mean salary and

standard deviation of salaries?Mean = 24 000 + 500 = $24 500Standard deviation = $600

Question 7

The mean salary at Carter-Drumfield sportswear store is $24 000 per annum with a standard deviation of $600. All employees get a $500 rise per annum.b. If instead of a flat rise, each employee had an income increase of 2%, what would be the new mean salary and standard deviation?

Question 7

The mean salary at Carter-Drumfield sportswear store is $24 000 per annum with a standard deviation of $600. All employees get a $500 rise per annum.b. If instead of a flat rise, each employee had an income increase of 2%, what would be the new mean salary and standard deviation?

Question 8

Adult males of a certain species are known to have a mean weight of 1.3 kg with a standard deviation of 0.2 kg. Adult females have a mean weight of 0.9 kg with a standard deviation of 0.1 kg.a. What is the mean weight and standard deviation of randomly selected groups of 2 males and 4 females?b. What is the probability that such a group will have a weight exceeding 6.5 kg? State your assumptions.

Question 8

Adult males of a certain species are known to have a mean weight of 1.3 kg with a standard deviation of 0.2 kg. Adult females have a mean weight of 0.9 kg with a standard deviation of 0.1 kg.a. What is the mean weight and standard deviation of

randomly selected groups of 2 males and 4 females?

Question 8

Adult males of a certain species are known to have a mean weight of 1.3 kg with a standard deviation of 0.2 kg. Adult females have a mean weight of 0.9 kg with a standard deviation of 0.1 kg.a. What is the mean weight and standard deviation of

randomly selected groups of 2 males and 4 females?

Question 8

Adult males of a certain species are known to have a mean weight of 1.3 kg with a standard deviation of 0.2 kg. Adult females have a mean weight of 0.9 kg with a standard deviation of 0.1 kg.b. What is the probability that such a group will have a weight exceeding 6.5 kg? State your assumptions.

Question 8

Adult males of a certain species are known to have a mean weight of 1.3 kg with a standard deviation of 0.2 kg. Adult females have a mean weight of 0.9 kg with a standard deviation of 0.1 kg.b. What is the probability that such a group will have a

weight exceeding 6.5 kg? State your assumptions.We assume that the weights are normally distributed.Mean = 6.2SD = 0.346

Question 8

b. What is the probability that such a group will have a weight exceeding 6.5 kg? State your assumptions.

We assume that the weights are normally distributed.Mean = 6.2SD = 0.346

6.2 6.5

Question 9

Two bags have black and white counters. Bag 1: 3 black and 1 white Bag 2: 6 black and 2 white.a. Which bag gives a better chance of picking a black counter?b. Which bag gives a better chance of picking two black counters?

Question 9

Two bags have black and white counters. Bag 1: 3 black and 1 white Bag 2: 6 black and 2 white.a. Which bag gives a better chance of picking a

black counter?There is an equal chance

Question 9

Two bags have black and white counters. Bag 1: 3 black and 1 white Bag 2: 6 black and 2 white.b. Which bag gives a better chance of picking

two black counters?

Question 10

Assume that the chance of having a boy or girl is the same. Over the course of a year, in which type of hospital would you expect there to be more days on which at least 60% of the babies born were boys? Give reasons for your answer.a. In a large hospitalb. In a small hospitalc. It makes no difference

Question 10

Assume that the chance of having a boy or girl is the same. Over the course of a year, in which type of hospital would you expect there to be more days on which at least 60% of the babies born were boys? Give reasons for your answer.a. In a large hospitalb. In a small hospitalc. It makes no difference

Question 11

• A cab was involved in a hit-and-run accident at night. There are two cab companies that operate in the city, a Blue Cab company and a Green Cab company. It is known that 85% of the cabs in the city are Green and 15% are Blue. A witness at the scene identified the cab involved in the accident as a Blue Cab. The witness was tested under similar visibility conditions, and made correct colour identifications in 80% of the trial instances. What is the probability that the cab involved in the accident was a Blue cab as stated by the witness?

• If your answer was 80%, you are in the majority.

• The 80% answer shows how we have a tendency to primarily consider only the last evidence given to us, ignoring earlier evidence.

• If we are simply told that a cab was involved in a hit-and-run accident, and are not given the information about the witness, then the majority of us will correctly estimate the probability of it being a Blue cab as 15%.

• Given new evidence (the 80% reliable witness) we throw away the first calculation and base our answer solely on the reliability of the witness. We do this to simplify the calculation, but in this case it leads to the wrong answer.

There are four possible scenarios.

• Green (85%) and correctly identified as Green (80%). Chance is 68%

• Green (85%) and misidentified as Blue (20%). Chance is 17%

• Blue (15%) and correctly identified as Blue (80%). Chance is 12%

• Blue (15%) and misidentified as Green (20%). Chance is 3%

Conditional probability

• In this case, we know that the witness said it was a Blue cab, so we only need to consider those cases where the cab was identified as Blue.

• That means it was either a misidentified Green (17%) or a correctly identified Blue (12%). So the chance that it was actually Blue is the chance of it being correctly identified as Blue (12%) over the chance that it was identified as Blue, whichever colour it actually was (12% + 17%, or 29%). That means that the chance of it being Blue, after being identified as Blue, is 12/29, or about 41%.

• The chance that it was actually Green is the remaining 59%.

• But with a witness who is 80% reliable, how can he be so likely to get it wrong? The catch is that the small chance of his incorrect identification is swamped by the huge number of Green cabs, which just make it so much more likely that any cab in the city is Green.

• Basically, with a compound probability like this you have to be careful to check out the contribution of both the correct (correctly identified Blue) and the incorrect (misidentified Green) terms. Otherwise, you may miss a large contribution which works against your intuition.

• If only 5% of the cabs in the city are Blue, the chances drop to 4/23, or 17%. In other words, if only 5% of the cabs are Blue and our 80% reliable witness identifies a Blue cab in an accident, there is only a 17% chance that he's actually right. Our 80% reliable witness is 5 times more likely to be wrong than right!