Binomial Distribution

Binomial -> 2 outocmes {Success or Fail}

n trials

independent trials

Same Probabilty of success throughotu the trial

/Sampling without replacement from a large population:

(trials can be considered independent as long as the sample size is less than 10% of the population size.)/

n < 10% of N

[1 million 90% soccer]

take one from this and the probability of the first is 900,000/1,000,000 1st person

take another from this and the probability is 899,999/1,000,000 ~ .8999 ~ .9 2nd person

Two electronic components are selected from the production line of 100 such components for thorough inspection. It is known that 90% of the components are “good”. Find the probability that the two inspected components are “good”.

2 outcomes (good or bad)

n = 2

2/100 = 2%<10% -> independant

p = .9%

Binomial [x]

Probability of both being good

P(both good)

A = 1st good B = 2nd good

P(AnB) = P(A)*P(B|A)

(.9)(.9) = .81

technically it is (.9)(.89) = .8091

but because its binomial we can assume independance!

At 25oC, 80% of a certain type of laser diodes have efficiency above 0.3 mW/mA. For five diodes, selected by simple random sampling from a large population of such diodes, find the probability of the following events.

Success = S = efficiency above .3…

Failure = F = efficiency below .3…

n = 5

5 from a large population so lets assume its less than 10%

let X = {# successes}

p = .8 in each trial

Binomial [x]

find probability of..

1. all five have efficiency above .3 at 25C

   .8⁵ = .32768

   So, P(X = 5) = .328

2. only the second diode selected has efficiency below .3 at 25C

   –> SFSSS

   Cant have P(x= 4) because F can be anyware

   (.2)*(.8⁴) = .08192

3. Exactly one of the five diodes has effiency below .3 at 25C

   –> e.g. FSSSS, SFSSS, … , 5 total

   5*(.2)*(.8⁴)

   So, P(x = 4) = .4096

4. exactly two of the five diodes have effiency below .3 at 25C

   –> e.g. FSFSS, SFFSS, P1P2 = P2P1 (Order does not matter)

   (5*4)/2! = 10 ways

   10*.2²*.8³ = .2048

Suppose 70% of all purchases in a certain store are made with credit card. Let X denote the number of credit card uses in the next 10 purchases. (Suppose there is a large population of purchases made at this store.)

S = {credit, no credit}

x = (# of credit card uses)

n = 10 purchases

P = .7

Assume independance because large population

Binomial [x]

Find P(5 ≤ X ≤ 8).

p(x=5) + p(x=6) + p(x=7) + p(x=8)

P(x=5)

SSSSSFFFFF -> .7⁵ * .3⁵

(10 choose 5) * .7⁵ * .3⁵

P(x=6)

SSSSSSFFFF -> .7⁶ * .3⁴

(10 choose 6) * .7⁶ * .3⁴

P(x=7)

SSSSSSSFFF -> .7⁷ * .3³

(10 choose 7) * .7⁷ * .3³

P(x=8)

SSSSSSSFFF -> .7⁸ * .3²

(10 choose 8) * .7⁸ * .3²

Equation

&Sigma; r = 5 to 8 (10 choose r) .7^r*3^(10-r)