8th Edition

ISBN: 9781337904285

Author: Larson

Publisher: CENGAGE L

Concept explainers

It simply refers to the likelihood of an event taking place whenever the occurrence of an event is uncertain. The probability of a single event can be calculated by dividing the number of successful trials of that event by the total number of trials.

Expected Value

When a large number of trials are performed for any random variable ‘X’, the predicted result is most likely the mean of all the outcomes for the random variable and it is known as expected value also known as expectation. The expected value, also known …

Probability Distributions

Understanding probability is necessary to know the probability distributions. In statistics, probability is how the uncertainty of an event is measured. This event can be anything. The most common examples include tossing a coin, rolling a die, or choosi…

Basic Probability

The simple definition of probability it is a chance of the occurrence of an event. It is defined in numerical form and the probability value is between 0 to 1. The probability value 0 indicates that there is no chance of that event occurring and the prob…

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Question

Chapter 8.6, Problem 70E

To determine

To explain: why the following given pattern gives the probability that the n people have distinct birthdays.

Expert Solution

Answer to Problem 70E

The probability of no two people from the group of n people sharing the same birthday is

$P(E)=n(E)n(S)=365·364·363.....[365−(n−1)]365n.$

Explanation of Solution

Given information: Consider a group of n people, the given pattern is,

$n=2;365365·364365=365·3643652n=3;365365·364365·363365=365·364·3633653$

Calculation:

The pattern gives the probability that n people have distinct birthdays using the standard formula of $P(E)=n(E)n(S)$ . The sample space is the total number of possible combinations of birthdays amongst the n people. Since each person could be born on any of the 365 days of the year, this number is $356n$ . In order for all these people to have different birthdays, the first person could be born on any of the 365 days of the year, the second on any of the 365 days of the year on which the first was not born, the third on any of the 365 days the first two were not born on and so on. Therefore the probability of no two people from the group of n people sharing the same birthday is

$P(E)=n(E)n(S)=365·364·363.....[365−(n−1)]365n.$

To determine

To write: an expression for the probability that four people ( n =4) have distinct birthdays using the pattern in part (a).

Expert Solution

Answer to Problem 70E

$n=4 ; 365365·364365·363365·362365=365·364·363·3623654$

Explanation of Solution

Given information: Consider a group of n people, the given pattern is,

$n=2;365365·364365=365·3643652n=3;365365·364365·363365=365·364·3633653$

Calculation:

To find the probability that amongst 4 people , no 2 share the same birthday, plug n= 4 into the given pattern.

$n=4 ; 365365·364365·363365·362365=365·364·363·3623654$

To determine

To verify: that $Pn$ probability that the n people have distinct birthdays can be obtained recursively by

$P1=1 and Pn=365−(n−1)365Pn−1.$

Expert Solution

Answer to Problem 70E

$Pn$ Probability that the n people have distinct birthdays can be obtained recursively by

$P1=1 and Pn=365−(n−1)365Pn−1.$

Explanation of Solution

Given information: Given $Pn$ probability that the n people have distinct birthdays can be obtained recursively by

$P1=1 and Pn=365−(n−1)365Pn−1.$

Calculation:

Here, use the probability of 4 people having different birthdays as an example of how the recursive method provided is a valid way to find this probability. Using the recursive definition,

$P4=365−(4−1)356P4−1=362356P3=362356·365−(3−1)356P3−1=363·3623562P2=363·3623562·365−(2−1)356P2−1=364·363·3623563P1=364·363·3623563·365365=365·364·363·3623564.$

Same result using this recursive definition as the one obtained in part (b).

Hence, Verified.

To determine

To explain: why $Qn=1−Pn$ gives the probability that at least two people in a group of n people have the same birthday.

Expert Solution

Answer to Problem 70E

$Qn=1−Pn$ .

Explanation of Solution

Given information: $Qn=1−Pn$ gives the probability that at least two people in a group of n people have the same birthday.

Calculation:

$Pn$ gives the probability that in a group of n people , none share the same birthday. The complement of the event is $Qn$ , is the event that at least two people share the same birthday. Therefore, by the law of complementation, $Qn=1−Pn .$

To determine

To complete: the given table using the result of parts (c) and (d).

Expert Solution

Answer to Problem 70E

n	10	15	20	23	30	40	50
$Pn$	0.88	0.75	0.59	0.49	0.29	0.11	0.03
$Qn$	0.12	0.25	0.41	0.51	0.71	0.89	0.97

Explanation of Solution

Given information: Given table is,

n	10	15	20	23	30	40	50
$Pn$
$Qn$

Calculation:

When constructing the probability of this table, it can help to recognize that:

$Pn=365·364·.....[365−(n−1)]365n=365!365n(365−n)!=365Pn365n.Qn=1−Pn$

The completed table shown below,

n	10	15	20	23	30	40	50
$Pn$	0.88	0.75	0.59	0.49	0.29	0.11	0.03
$Qn$	0.12	0.25	0.41	0.51	0.71	0.89	0.97

To determine

To find: how many people must be in a group so that the probability of at least two of them having the same birthday is greater than $12$ , explain.

Expert Solution

Answer to Problem 70E

23 people in a group so that the probability of at least two of them having the same birthday is greater than $12$ , explain.

Explanation of Solution

Given information:

n	10	15	20	23	30	40	50
$Pn$	0.88	0.75	0.59	0.49	0.29	0.11	0.03
$Qn$	0.12	0.25	0.41	0.51	0.71	0.89	0.97

Calculation:

It is clear from the above table, 23 people must be in a group for the probability of at least two them having the same birthday to be greater than $12.$

Chapter 8.6, Problem 69E

Chapter 8.6, Problem 71E

Chapter 8 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E Ch. 8.1 - Prob. 8E Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E

Ch. 8.1 - Prob. 11E Ch. 8.1 - Prob. 12E Ch. 8.1 - Prob. 13E Ch. 8.1 - Prob. 14E Ch. 8.1 - Prob. 15E Ch. 8.1 - Prob. 16E Ch. 8.1 - Prob. 17E Ch. 8.1 - Prob. 18E Ch. 8.1 - Prob. 19E Ch. 8.1 - Prob. 20E Ch. 8.1 - Prob. 21E Ch. 8.1 - Prob. 22E Ch. 8.1 - Prob. 23E Ch. 8.1 - Prob. 24E Ch. 8.1 - Prob. 25E Ch. 8.1 - Prob. 26E Ch. 8.1 - Prob. 27E Ch. 8.1 - Prob. 28E Ch. 8.1 - Prob. 29E Ch. 8.1 - Prob. 30E Ch. 8.1 - Prob. 31E Ch. 8.1 - Prob. 32E Ch. 8.1 - Prob. 33E Ch. 8.1 - Prob. 34E Ch. 8.1 - Prob. 35E Ch. 8.1 - Prob. 36E Ch. 8.1 - Prob. 37E Ch. 8.1 - Prob. 38E Ch. 8.1 - Prob. 39E Ch. 8.1 - Prob. 40E Ch. 8.1 - Prob. 41E Ch. 8.1 - Prob. 42E Ch. 8.1 - Prob. 43E Ch. 8.1 - Prob. 44E Ch. 8.1 - Prob. 45E Ch. 8.1 - Prob. 46E Ch. 8.1 - Prob. 47E Ch. 8.1 - Prob. 48E Ch. 8.1 - Prob. 49E Ch. 8.1 - Prob. 50E Ch. 8.1 - Prob. 51E Ch. 8.1 - Prob. 52E Ch. 8.1 - Prob. 53E Ch. 8.1 - Prob. 54E Ch. 8.1 - Prob. 55E Ch. 8.1 - Prob. 56E Ch. 8.1 - Prob. 57E Ch. 8.1 - Prob. 58E Ch. 8.1 - Prob. 59E Ch. 8.1 - Prob. 60E Ch. 8.1 - 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why the following given pattern gives the probability that the n people have distinct birthdays.

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To explain: why the following given pattern gives the probability that the n people have distinct birthdays.

Answer to Problem 70E

Explanation of Solution

To write: an expression for the probability that four people ( n =4) have distinct birthdays using the pattern in part (a).

Answer to Problem 70E

Explanation of Solution

Answer to Problem 70E

Explanation of Solution

To explain: why Qn=1−Pn<m:math><m:mrow><m:msub><m:mi>Q</m:mi><m:mi>n</m:mi></m:msub><m:mo>=</m:mo><m:mn>1</m:mn><m:mo>−</m:mo><m:msub><m:mi>P</m:mi><m:mi>n</m:mi></m:msub></m:mrow></m:math> gives the probability that at least two people in a group of n people have the same birthday.

Answer to Problem 70E

Explanation of Solution

To complete: the given table using the result of parts (c) and (d).

Answer to Problem 70E

Explanation of Solution

To find: how many people must be in a group so that the probability of at least two of them having the same birthday is greater than 12<m:math><m:mrow><m:mfrac><m:mn>1</m:mn><m:mn>2</m:mn></m:mfrac></m:mrow></m:math> , explain.

Answer to Problem 70E

Explanation of Solution

Chapter 8 Solutions

To explain: why $Qn=1−Pn$ gives the probability that at least two people in a group of n people have the same birthday.

To find: how many people must be in a group so that the probability of at least two of them having the same birthday is greater than $12$ , explain.