Hypergeometric Distributions When choosing the starting line-up for a game, a

coach has to select a different player for each starting position – obviously!

Similarly, when a person is elected to represent student council or you are dealt a card from a standard deck, there can be no repetition.

In such situations, each selection reduces the number of items that could be selected in the next trial.

Thus, the probabilities in these trials are dependent Often, we need to compute the probability of a specific

number of successes in a given number of dependent trials.

Hypergeometric Distributions Recall, the concepts associated with a

Binomial Distribution: n identical trials (Bernouli Trials) Two possible outcomes (Success or Failure) Probability of Success does not change with each

trial Trials are independent of one another

Purpose is to determine how many successes occur in n trials.

Hypergeometric Distributions With Hypergeometric Distributions, some of

the concepts are transferable… Each trial has possibility only for success or failure Specific number of trials Random variable is the number of successful

outcomes from the specified number of trials Individual outcomes cannot be repeated within the

trials

Hypergeometric Distributions There are also some differences…

Each trial is dependent on the previous trial The probability of success changes with

each trial Calculations of probabilities in

Hypergeometric Distributions generally require formulae using combinations

Hypergeometric DistributionsExample 1 Civil trials in Ontario require 6 jury

members. Suppose a civil-court jury is being selected from a pool of 18 citizens, 8 of whom are men.

a) Determine the probability distribution for the number of women on the jury.

b) What is the expected number of women on the jury?

Hypergeometric DistributionsSol’n Selection process involves dependent events

since each person who gets chosen cannot be selected again.

Look to combinations for total number of ways 6 jurors can be selected from the pool of 18…

564,186

18)(

Sn

Hypergeometric DistributionsCont… There can be between 0 and 6 women on the

jury. The number of ways in which x women can be selected is 10Cx.

Thus, the men can fill the remaining 6-x positions on the jury in 8C6-x ways.

Therefore, the number of ways of selecting a jury with x women on it is the product of the two…

Hypergeometric DistributionsCont… Probability of a jury with x women is…

6

18

6

810

)(

)()(

xx

Sn

xnxP

Hypergeometric DistributionsCont… Using the formula for

the probability…

# of women x

Probability

P(x)

0 0.00151

1 0.03017

2 0.16968

3 0.36199

4 0.31674

5 0.10860

6 0.01131

6

18

6

810

)(xx

xP

Hypergeometric DistributionsCont…

Graphically…Probability of Women on Jury

00.050.1

0.150.2

0.250.3

0.350.4

0 1 2 3 4 5 6

Number of women (X)

Prob

abili

ty

Hypergeometric DistributionsCont…

b) Expected Value…

n

iii xPxxE

0

)()(

333.3

01131.0610860.0531674.04

36199.0316968.0203017.0100151.00

Hypergeometric Distributions Generalizing the methods lead to…Probability in a Hypergeometric DistributionThe probability of x successes in r dependent trials

-where: n population size & a is the number of successes in the population

r

n

xr

an

x

a

xP

Hypergeometric DistributionsExpected Value for a Hypergeometric Distribution

Notes: Ensure that the number of trials is representative of the

situation Each trial is dependent (no replacement between trials)

n

raxE )(

Hypergeometric DistributionsExample 2

A box contains seven yellow, three green, five purple, and six red candies jumbled together.

1. What is the expected number of red candies among five candies poured from the box?

2. Verify that the expected value formula for H.D. gives the same value as the expectation for any probability distribution.

Hypergeometric DistributionsSol’n

1. n=7+3+5+6=21(# of candies in box {popl’n})

r=5 (# of candies removed {trials})

a=6 (# of red candies {successes})

4285.121

65

)(

n

raxE

Hypergeometric DistributionsCon’t…

2. Using the general expectation formula

)()( xxPxE

4285.1

5

21

0

15

5

6

5

5

21

1

15

4

6

4

5

21

2

15

3

6

3

5

21

3

15

2

6

2

5

21

4

15

1

6

1

5

21

5

15

0

6

0

Hypergeometric DistributionsExample 3

In wildlife management, the MoE caught and tagged 500 raccoons in a wilderness area. The raccoons were released after being vaccinated against rabies. To estimate the raccoon population in the area, the ministry caught 40 raccoons during the summer. Of these, 15 had tags.

1. Discuss why this can be modeled with a hypergeometric distribution.

2. Estimate the raccoon population in the area.

Hypergeometric Distributions1. The 40 raccoons captured in the summer

were all different from each other. In other words, there were no repetitions, thus the trials were dependent. The captured raccoon was either tagged (success) or not (failure). Therefore, the situation has the characteristics of a hypergeometric distribution.

Hypergeometric Distributions2. Assume that the number of tagged

raccoons caught in the summer is equal to the expected number of raccoons for the hypergeometric distribution. Substitute the the known values into the formula and solve for the population size, n.

r=40 (# raccoons caught in summer {trials})

a=500 (# tagged raccoons {population})

Hypergeometric Distributions

So,

Therefore, the # of raccoons in the area is about 1333.

n

raxE )(

3.133315

50040

5004015

n

n

n

Hypergeometric DistributionsProbability in a Hypergeometric Distribution

The probability of x successes in r dependent trials is:

-where: n population size & a is the number of successes in the population

r

n

xr

an

x

a

xPn

raXE )(