Question

Rework problem 4 from section 4.1 of your text involving the computation of probabilities for Bernoulli trials. Use the following values instead of those found in your book.

(1) 3 successes in 5 trials with q = 0.7:  

(2) 3 failures in 6 trials with q = 0.6:

Answer 1

Using binomial probability calculator.

P( X=x) = (n C x)* (p^X)*(1-p)^(n-x)

(1) 3 successes in 5 trials with q = 0.7

Ans : let X ~ Binomial ( n = 5, p = 0.3)

P( X = 3) = ( 5 C 3) * (p ^ 3) *(1-p)^ 2

P( x= 3) = 0.1323

p(3 successes in 5 trials ) = 0.1323

(2) 3 failures in 6 trials with q = 0.6:

Ans : let X ~ binomial (n= 6,p = 0.4)

Where, p = 1-q

3 trials are fail out of 6 means 3 trila are success

Therefore,

P( X = 3 ) = (6 C 3)* (p ^ 3) * (1-p) ^3

P(x = 3 ) = 0.2765

Therefore ,

P(3 failures in 6 trials ) = 0.2765

Thank you!

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