Exercise 17 Page 716

To model the given situation, we want to use a binomial distribution. We have 5 repeated independent trials — each attempted free throw — and each trial has only two possible outcomes, success or failure. Since this basketball player makes their free throws 82.6 % of the time, we know that this is the probability of success p. Probability of Success p= 0.826 The sum of the probability of success and the probability of failure is always 1 in a binomial experiment. We are able to find the probability of failure q using this fact. Since we want to draw a histogram of the binomial distribution of the number of successful free throws, we need to remember the formula for the probability of exactly k successes in n trials for a binomial experiment. P(k successes)= _nC_k p^k q^(n- k) The player attempts 5 throws, so n= 5. Therefore, the number of possible successes varies from 0 to 5. We have all the necessary information to calculate any of the probabilities. Let's start with P( k= 0).

P(k=0) = _nC_k p^k q^(n-k)

SubstituteValues

Substitute values

P(k=0)= _5C_0 ( 0.826 )^0 ( 0.174 )^(5- 0)

▼

Simplify

SubTerm

Subtract term

P(k=0)= _5C_0 ( 0.826 )^0 ( 0.174 )^5

ExponentZero

a^0=1

P(k=0)= _5C_0 (1) ( 0.174 )^5

IdPropMult

Identity Property of Multiplication

P(k=0)= _5C_0 ( 0.174 )^5

Next, let's recall the formula to calculate _nC_k. _nC_k =n!/k!( n- k)! ⇒ _5C_0= 5! 0!( 5- 0)! We can substitute this into the expression on the right-hand side of our equation.

P(k=0)= _5C_0 ( 0.174 )^5

Substitute

_5C_0= 5!/0!( 5- 0)!

P(k=0)=5!/0!( 5- 0)! ( 0.174 )^5

▼

Simplify

SubTerm

Subtract term

P(k=0)=5!/0!*5! ( 0.174 )^5

ReduceFrac

a/b=.a /5!./.b /5!.

P(k=0)=1/0! ( 0.174 )^5

0!=1

P(k=0)=1/1 ( 0.174 )^5

QuotOne

a/a=1

P(k=0)=1 ( 0.174 )^5

IdPropMult

Identity Property of Multiplication

P(k=0)=( 0.174 )^5

UseCalc

Use a calculator

P(k=0) = 0.000159...

RoundDec

Round to 4 decimal place(s)

P(k=0) ≈ 0.0002

We can find the probabilities for other values of k in the same way.

Formula	Simplify	Probability
P(k=0) = _5C_0 (0.826)^0 (0.174)^(5-0)	P(k=0) = (0.174)^5	≈ 0.0002
P(k=1) = _5C_1 (0.826)^1 (0.174)^(5-1)	P(k=1) = 5 (0.826)(0.174)^4	≈ 0.0038
P(k=2) = _5C_2 (0.826)^2 (0.174)^(5-2)	P(k=2) = 10 (0.826)^2 (0.174)^3	≈ 0.0359
P(k=3) = _5C_3 (0.826)^3 (0.174)^(5-3)	P(k=3) = 10 (0.826)^3 (0.174)^2	≈ 0.1706
P(k=4) = _5C_4 (0.826)^4 (0.174)^(5-4)	P(k=4) = 5 (0.826)^4 (0.174)	≈ 0.4050
P(k=5) = _5C_5 (0.826)^5 (0.174)^(5-5)	P(k=5) = (0.826)^5	≈ 0.3845

We are ready to construct the histogram. The intervals are given by k and the frequencies are given by the calculated probabilities. Keep in mind that the bars should touch but not overlap. Do not forget to label the axes!

Analyzing the table and the histogram, we found that the most likely outcome is that the basketball player will make 4 successful free throws.