Math  /  Data & Statistics

QuestionJR makes 88%88 \% of all three-point shots and 85%85 \% of all free-throw shots while playing basketball. Suppose she shoots 8 three-point shots and 5 free-throw shows. What is the probability that she makes 5 three-point shots and 3 free-throw shots?

Studdy Solution

STEP 1

What is this asking? What's the chance JR sinks exactly 5 out of 8 three-pointers *and* 3 out of 5 free throws? Watch out! Don't mix up "and" and "or" probabilities!
We want both events to happen.

STEP 2

1. Three-Point Probability
2. Free-Throw Probability
3. Combined Probability

STEP 3

We're dealing with **independent events** here, meaning one shot doesn't affect another.
We want the probability of making *exactly* **5** out of **8** three-pointers.
This is a classic **binomial probability** problem!

STEP 4

The **binomial probability formula** is our friend: P(X=k)=(nk)pk(1p)nkP(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} Where nn is the **number of trials** (shots), kk is the **number of successes** (made shots), and pp is the **probability of success** on a single trial.

STEP 5

Let's plug in our three-point values: n=8n = \textbf{8} (total shots), k=5k = \textbf{5} (made shots), and p=0.88p = \textbf{0.88} (success rate). P(X=5)=(85)(0.88)5(10.88)85P(X=5) = \binom{8}{5} \cdot (0.88)^5 \cdot (1-0.88)^{8-5}

STEP 6

Calculate the binomial coefficient: (85)=8!5!(85)!=8!5!3!=876321=56\binom{8}{5} = \frac{8!}{5! \cdot (8-5)!} = \frac{8!}{5! \cdot 3!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56

STEP 7

Now, let's calculate the probability: P(X=5)=56(0.88)5(0.12)3P(X=5) = 56 \cdot (0.88)^5 \cdot (0.12)^3 P(X=5)=560.52773191680.0017280.0532P(X=5) = 56 \cdot 0.5277319168 \cdot 0.001728 \approx 0.0532So, the probability of making exactly **5** out of **8** three-pointers is approximately **0.0532**.

STEP 8

Time for free throws!
Same binomial probability deal, but with different numbers.

STEP 9

Now, n=5n = \textbf{5} (total free throws), k=3k = \textbf{3} (made free throws), and p=0.85p = \textbf{0.85} (free throw success rate). P(Y=3)=(53)(0.85)3(10.85)53P(Y=3) = \binom{5}{3} \cdot (0.85)^3 \cdot (1-0.85)^{5-3}

STEP 10

Calculate the binomial coefficient: (53)=5!3!(53)!=5!3!2!=5421=10\binom{5}{3} = \frac{5!}{3! \cdot (5-3)!} = \frac{5!}{3! \cdot 2!} = \frac{5 \cdot 4}{2 \cdot 1} = 10

STEP 11

Calculate the probability: P(Y=3)=10(0.85)3(0.15)2P(Y=3) = 10 \cdot (0.85)^3 \cdot (0.15)^2 P(Y=3)=100.6141250.02250.1382P(Y=3) = 10 \cdot 0.614125 \cdot 0.0225 \approx 0.1382The probability of making exactly **3** out of **5** free throws is approximately **0.1382**.

STEP 12

Since the three-point shots and free throws are **independent events**, we **multiply** their probabilities to find the probability of *both* events happening.

STEP 13

P(X=5 and Y=3)=P(X=5)P(Y=3)P(X=5 \text{ and } Y=3) = P(X=5) \cdot P(Y=3) P(X=5 and Y=3)0.05320.13820.00735P(X=5 \text{ and } Y=3) \approx 0.0532 \cdot 0.1382 \approx 0.00735

STEP 14

The probability that JR makes exactly 5 three-point shots and 3 free-throw shots is approximately **0.00735**, or about **0.735%**.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord