Question

Let X be the number of successes in six independent trials of a binomial experiment in which the probability of success is p = 4 5 . Find the following probabilities. (Round your answers to four decimal places.)

(a) P(X = 5)

(b) P(3 ≤ X ≤ 5)

Answer 1

Let x = number of successes in six independent trials of a binomial experiment

p = probability of success is =0.45

n = sample size = n = 6

A) we want to calculate that P(X=5)

Binomial distribution has p.m.f. is

$P(X=x)=_{x}^{n}\textrm{C} * p^x *q^{n-x}$

For x = 5

$P(X=5)=_{5}^{6}\textrm{C} * 0.45^5 *(1-0.45)^{6-5}$

  $=6*0.1845*0.55$

$=0.0609$ ...............( Answer)

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B) we want to calculate that  P(3 ≤ X ≤ 5)

Binomial distribution has p.m.f. is

$P(X=x)=_{x}^{n}\textrm{C} * p^x *q^{n-x}$

By using Binomial distribution we can calculate it as

$P(3 \leq X \leq 5) = P(X=3)+P(X=4)+P(X=5)$

For x = 3

$P(X=3)=_{3}^{6}\textrm{C} * 0.45^3 *(1-0.45)^{6-3}$

  $=20*0.0911*0.1664$

  $=0.3032$

Similarly for x = 4

$P(X=4)=_{4}^{6}\textrm{C} * 0.45^4 *(1-0.45)^{6-4}$

  $=0.1861$

and for x = 5

$P(X=5)=_{5}^{6}\textrm{C} * 0.45^5 *(1-0.45)^{6-5}$

  $=0.0609$

we get $P(3 \leq X \leq 5) = P(X=3)+P(X=4)+P(X=5)$

  $=0.3032+0.1861+0.0609$

  $=0.5502$ ..................( ANSWER)