Probability and StatisticsProbability and Statistics

BINOMIALBINOMIAL

DISTRIBUTIONSDISTRIBUTIONS

Binomial DistributionBinomial Distribution

Binomial distribution is a suitable model for Binomial distribution is a suitable model for event which has:event which has:

(i) (i) Exactly 2 possible outcomes that are Exactly 2 possible outcomes that are mutually exclusivemutually exclusive

(ii)(ii) Fixed number of trialsFixed number of trials

(iii)(iii) Independent outcomesIndependent outcomes

(iv)(iv) Constant probability of successConstant probability of success

(v)(v) P(success) + P(failure) = 1P(success) + P(failure) = 1

FormulaFormula

If X follows Binomial distribution with If X follows Binomial distribution with nn number number of trials and of trials and pp, the probability of success then:, the probability of success then:

Where x = 0, 1, 2, ….., nWhere x = 0, 1, 2, ….., n

xnx.qnCx.px)P(X

Calculator key-inCalculator key-in

For a Binomial distribution with For a Binomial distribution with

n = n = number of trials and number of trials and

p = p = probability of success:probability of success:

On > 2On > 2ndnd and VARS > BINOMPDF and VARS > BINOMPDF

P(X = x) = binompdf(n, p, x)P(X = x) = binompdf(n, p, x)

Probability for Probability for one value one value onlyonly

Where x = 0, 1, 2, ….., nWhere x = 0, 1, 2, ….., n

Calculator key-inCalculator key-in

On > 2On > 2ndnd and VARS > BINOMCDF and VARS > BINOMCDF

P(X P(X x) = binomcdf(n, p, x) x) = binomcdf(n, p, x)

Probability for Probability for cumulative values cumulative values up to x,up to x, i.e., for values 0, 1, ...., xi.e., for values 0, 1, ...., x

Where x = 0, 1, 2, ….., nWhere x = 0, 1, 2, ….., n

Example 1Example 1

Joseph and four friends each have an Joseph and four friends each have an independent probability 0.45 of winning a independent probability 0.45 of winning a prize. Find the probability that:prize. Find the probability that:

(i) Exactly two of the five friends win a (i) Exactly two of the five friends win a prize.prize.

(ii) Joseph and only one friend win a prize.(ii) Joseph and only one friend win a prize.

SolutionSolution....

Here n = 5 (number of friends)Here n = 5 (number of friends)and p = 0.45 (probability of winning)and p = 0.45 (probability of winning)(i) P (Exactly two win) (i) P (Exactly two win)

= = 55CC22 0.45 0.4522 0.55 0.553 3 == 0.3370.337

binompdf(5, .45, 2) = .337binompdf(5, .45, 2) = .337(ii) P (Joseph and one friend win) (ii) P (Joseph and one friend win)

= 0.45 = 0.45 44CC11 0.45 0.45 0.55 0.553 3 == 0.1350.135

0.45 binompdf(4, .45, 1) = .1350.45 binompdf(4, .45, 1) = .135

Example 2Example 2

Eggs produced at a farm are packaged in boxes Eggs produced at a farm are packaged in boxes of six. For any egg, the probability that it is of six. For any egg, the probability that it is broken when it reaches the retail outlet is 0.1, broken when it reaches the retail outlet is 0.1, independent of other eggs. A box is said to be independent of other eggs. A box is said to be bad if it contains at least two broken eggs.bad if it contains at least two broken eggs.

(i) Calculate the probability that a randomly (i) Calculate the probability that a randomly selected box is bad.selected box is bad.

(ii) Ten boxes are chosen at a time, find the (ii) Ten boxes are chosen at a time, find the probability that just two of these boxes are bad.probability that just two of these boxes are bad.

SolutionSolution

(i) Here n = 6 and p = 0.1 (probability that an egg (i) Here n = 6 and p = 0.1 (probability that an egg is broken).is broken).

P (box is bad) P (box is bad) = 1 – P( box is good)= 1 – P( box is good)= 1 – P (Less than 2 eggs are broken)= 1 – P (Less than 2 eggs are broken)= 1 - {P(0) + P(1)}= 1 - {P(0) + P(1)}= 1 – {= 1 – {66CC00 0.1 0.100 0.9 0.96 6 + + 66CC11 0.1 0.1 0.9 0.955}}

= 1 – {0.53144 + 0.35429}= 1 – {0.53144 + 0.35429}= 0.114= 0.1141 – binomcdf(6, .1, 1) = .1141 – binomcdf(6, .1, 1) = .114

SolutionSolution..

(i) Here n = 10 (number of boxes)(i) Here n = 10 (number of boxes)

and p = 0.114 (probability that a box and p = 0.114 (probability that a box is bad).is bad).

P (two boxes are bad) P (two boxes are bad)

= = 1010CC22 0.114 0.11422 0.886 0.8868 8

= 0.222= 0.222

VarianceVariance

If X follows Binomial distribution with ‘If X follows Binomial distribution with ‘nn’ ’ number of trials and ‘number of trials and ‘pp’, probability of ’, probability of success then:success then:

Mean = Mean = = np = np

Variance = Variance = 22 = npq = npq

Example 3Example 3

ExerciseExercise