Exercise 39 Page 745

If we have n repeated independent trials, each with a probability of success p and a probability of failure q, then the Binomial Probability of x successes in the n trials can be found by the following formula. P(x)= _nC_x p^x q^(n-x) We are told that p= 0.4. With the value of p and knowing that p+q=1, we can calculate the value of q.

p+q=1

Substitute

p= 0.4

0.4+q=1

SubEqn

LHS-0.4=RHS-0.4

q= 0.6

We are also told that x= 2 and n= 9. We can substitute all of these values into the formula and evaluate the right-hand side.

P(x)= _nC_x p^x q^(n-x)

SubstituteValues

Substitute values

P( 2)= _9C_2 ( 0.4)^2 ( 0.6)^(9- 2)

Next, let's recall the formula to calculate _nC_x. _nC_x =n!/x!(n-x)! Therefore, we can substitute 9!2!(9-2)! for _9C_2.

P(2)= _9C_2 (0.4)^2 (0.6)^(9-2)

Substitute

_9C_2= 9!/2!(9-2)!

P(2)= 9!/2!(9-2)! (0.4)^2 (0.6)^(9-2)

▼

Evaluate right-hand side

SubTerms

Subtract terms

P(2)=9!/2!(7)! (0.4)^2 (0.6)^7

Write as a product

P(2)= 9(8)(7!)/2(1)(7!) (0.4)^2 (0.6)^7

CancelCommonFac

Cancel out common factors

P(2)= 9(8)(7!)/2(1)(7!) (0.4)^2 (0.6)^7

SimpQuot

Simplify quotient

P(2)= 9(8)/2(1) (0.4)^2 (0.6)^7

Multiply

Multiply

P(2)= 72/2 (0.4)^2 (0.6)^7

CalcQuot

Calculate quotient

P(2)=36 (0.4)^2 (0.6)^7

CalcPow

Calculate power

P(2)=36 (0.16) (0.0279936)

Multiply

Multiply

P(2)=0.161243

RoundDec

Round to 4 decimal place(s)

P(2)≈ 0.1612

If the probability of success for each trial is 0.4, then the probability of 2 successes in 9 trials is about 0.1612.