Exercise 40 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= 0.3. With the value of p and knowing that p+q=1, we can calculate the value of q.

p+q=1

Substitute

p= 0.3

0.3+q=1

SubEqn

LHS-0.3=RHS-0.3

q= 0.7

We are also told that x= 6 and n= 10. 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( 6)= _(10)C_6 ( 0.3)^6 ( 0.7)^(10- 6)

Next, let's recall the formula to calculate _nC_x. _nC_x =n!/x!(n-x)! Therefore, we can substitute 10!6!(10-6)! for _(10)C_6.

P(6)= _(10)C_6 (0.3)^6 (0.7)^(10-6)

Substitute

_(10)C_6= 10!/6!(10-6)!

P(6)= 10!/6!(10-6)! (0.3)^6 (0.7)^(10-6)

▼

Evaluate right-hand side

SubTerms

Subtract terms

P(6)= 10!/6!(4!) (0.3)^6 (0.7)^4

Write as a product

P(6)= 10(9)(8)(7)(6!)/6!(4)(3)(2)(1) (0.3)^6 (0.7)^4

CancelCommonFac

Cancel out common factors

P(6)= 10(9)(8)(7)(6!)/6!(4)(3)(2)(1) (0.3)^6 (0.7)^4

SimpQuot

Simplify quotient

P(6)= 10(9)(8)(7)/4(3)(2)(1) (0.3)^6 (0.7)^4

Multiply

Multiply

P(6)= 5040/24 (0.3)^6 (0.7)^4

CalcQuot

Calculate quotient

P(6)=210 (0.3)^6 (0.7)^4

CalcPow

Calculate power

P(6)= 210 (0.000729) (0.2401)

Multiply

Multiply

P(6)=0.036757...

RoundDec

Round to 4 decimal place(s)

P(6)≈ 0.0368

If the probability of success for each trial is 0.3, then the probability of 6 successes in 10 trials is about 0.0368.