many experiments can be done with the results of each trial reduced to 2 outcomes binomial...

Many Experiments can be done with the results of each trial reduced to 2

outcomes• Binomial Experiment:

• There are n independent trials

• Each trial has only 2 possible outcomes:

• Success or failure.

• The probability of success is the same for each trial.

• The probability is p.

• The probability of failure is 1 – p

Finding a Binomial Probability• For a binomial experiment consisting of n trials,

the probability of exactly K successes is:

• P(k successes) = nCk pk (1 – p)n - k

• Where the probability of success on each trial is p.

EXAMPLE 1• At a college, 53% of students receive financial

aid. In a random group of 9 students, what is the probability that exactly 5 of them receive financial aid?

• p=.53 (the prob of success for each trial)

• n=9 (diff trials or experiments)

• The prob of getting 5 successes (k=5)

• P(k=5) = 9C5 .535 (1-.53)9-5

• about 26%

Binomial Probability Theorem

The binomial theorem is used to calculate the probability for the outcomes of repeated independent and identical trials. If p is the probability of success and q is the probability of failure (q = 1 - p), then the probability of k successes in n trials is:

P(k successes) = nCk pk qn - k

2. Hockey cards, chosen at random from a set of 20, aregiven away inside cereal boxes. Stan needs one more card to complete his set so he buys five boxes of cereal. What is the probability that he will complete his set?

5 (number of trials)

1 (number of successes)

1 (probability of success)

2019

(probability of failure)20

n

k

p

q

P(x successes) = nCk pk qn - k

P(1 success) 5 C1

1

20

1

19

20

4

= 0.2

The probability of Stan completing his set is 20%.

3. Seven coins are tossed. What is the probability of four tails and three heads?


= 0.273

4. A true-false test has 12 questions. Suppose you guess all 12. What is the probability of exactly seven correct answers?


= 0.193

The probability of seven correct answers is 19%.

The probability of four heads and three tails is 27%.

Binomial Distribution - Applications

n = 7 (number of trials)

k = 4 (number of successes)

p = (probability of success)

q = (probability of failure)

1

21

2

P(4 successes) = 7 44 31

212

C ( ) ( )



p = (probability of success)1

2

q = (probability of failure)1

2

P(7 successes) = 12 77 51

212

C ( ) ( )

5. A test consists of 10 multiple choice questions, each with four possible answers. To pass the test, one must answer at least nine questions correctly. Find the probability of passing, if one were to guess the answer for each question.




43


n

k

p

q

P(k successes) = nCk px qn - k

P(at least 9 successes) = P(9 successes) + P(10 successes)




43


n

k

p

q

10C9

1

4

9

3

4

1

10C10 1

4

10

3

4

0

= 0.000 0296

The probability of passing is 0.003%.


7. A family has nine children. What is the probability that there is at least one girl?

This can be best solved using the compliment, that is, the probability of zero girls:


= 0.001 95

The probability of zero girls is 0.001 95, therefore the probability of at least one girl is 1 - 0.001 95 = 0.998.


9 00 91

212

C ( ) ( )P(0 successes) =



p = (probability of success)

n = (probability of failure)

1212


8. While pitching for the Toronto Blue Jays, 4 of every 7 pitches Juan Guzman threw in the first 5 innings were strikes. What is the probability that 3 of the next 4 pitches will be strikes?

n = 4 (number of pitches)

k = 3 (number of strikes)

p = (probability of success)7

4

q = (probability of failure)7

3


= 0.32

P(3 successes) =

7

3

7

43

34C

Let’s assume that there are only two possible outcomes, strikes or balls.

4 1CNote: will also yield the same result.

Using the Binomial Theorem to Calculate Probabilities

9. What is the probability of correctly guessing the outcome ofexactly one out of four rolls of a die?

The probability of correctly guessing one roll of the die is1

6.

The probability of incorrectly guessing the outcome is5

6.

The probability of one correct and three incorrect guesses is1

6

5

6

3

.

The correct guess can occur in any one of the four rolls so there are 4C1 ways of arranging the correct guess.

P(one correct guess in four rolls) = 4C1 1

6

1

5

6

3

= 0.386

NOTE: This experiment is called a binomial experiment because it has two outcomes: guessing correctly guessing incorrectly

Using the Binomial Theorem to Calculate Probabilities

For this experiment, let p represent the probability of a correct guess and q represent the probability of an incorrect guess.Use the binomial theorem. Expand and evaluate (p + q)6, where

1 5 and .

6 6p q

1

6

5

6

6

6C6

1

6

6 5

6

0

6C5

1

6

5 5

6

1

6C4

1

6

4 5

6

2

6C3

1

6

3 5

6

3

6C2

1

6

2 5

6

4

6C1

1

6

1 5

6

5

6C0

1

6

0 5

6

6

Thus, the probability of correctly guessing the outcome of six out of six rolls is

6

6 6

10.000021 4.

6C

10. Find the probability of correctly guessing the outcome ofexactly six out of six rolls of the die.NOTE: This is a binomial experiment because there are only two outcomes for each roll: guessing correctly or incorrectly.

Using the Binomial Theorem to Calculate Probabilities [cont’d]

You can use the table feature of a graphing calculator to calculate probabilities.

Using the Binomial Probability Distribution feature of the TI-83:

DISTR 0: binompdf

binompdf (number of trials, probability of success, k-value)

n = 6p = 0.17P(6 successes)

binompdf(6, 0.17, 6)= 0.000 021 4

The table shows theprobabilityfor any number ofcorrect guesses.

EXAMPLE 11• Draw a histogram of the binomial distribution

for the class of students in Example 1 and find the probability that fewer than 3 students in the class receive financial aid.

• Hint: Use P(k successes) = nCk pk (1-p)n-k

• (note: in a histogram, the bars should touch.)

EXAMPLE 1• At a college, 53% of students receive financial

aid. In a random group of 9 students, what is the probability that exactly 5 of them receive financial aid?

• p=.53 (the prob of success for each trial)

• n=9 (diff trials or experiments)

• The prob of getting 5 successes (k=5)

• P(k=5) = 9C5 .535 (1-.53)9-5

• about 26%

• P(k=0) = 9C0 .530 (1-.53)9-0 = .001• P(k=1) = 9C1 .531 (1-.53)9-1 = .011• P(k=2) = 9C2 .532 (1-.53)9-2 = .05• P(k=3) = 9C3 .533 (1-.53)9-3 = .13• P(k=4) = 9C4 .534 (1-.53)9-4 = .23• P(k=5) = 9C5 .535 (1-.53)9-5 = .26• P(k=6) = 9C6 .536 (1-.53)9-6 = .19• P(k=7) = 9C7 .537 (1-.53)9-7 = .09• P(k=8) = 9C8 .538 (1-.53)9-8 = .03• P(k=9) = 9C9 .539 (1-.53)9-9 = .003

Students on Financial Aid 53%

0.0010.0110.05

0.13

0.230.26

0.19

0.09

0.030.003

0

0.05

0.1

0.15

0.2

0.25

0.3

1 2 3 4 5 6 7 8 9 10

Students Surveyed

Pro

bab

ilit

y

• Probability of fewer than 3

• = P(0) + P(1) + P(2)

• = .001 + .011 + .05

• = 0.062

P. 742 #10-30

many experiments can be done with the results of each trial reduced to 2 outcomes binomial...

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