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Chapter 17: Problem 9

A basketball player has made \(80 \%\) of his foul shots during the season. Assuming the shots are independent, find the probability that in tonight's game he. a) misses for the first time on his fifth attempt. b) makes his first basket on his fourth shot. c) makes his first basket on one of his first 3 shots.

Short Answer

Expert verified
a) 0.08192, b) 0.0064, c) 0.992.

Step by step solution

01

Understand the Problem

We are given a probability problem involving a basketball player's shooting performance. His success rate is 80%, meaning the failure rate is 20%. We are tasked with finding probabilities for specific events related to his shooting in an upcoming game.
02

Use Geometric Distribution for Part (a)

The player misses a shot with a probability of 0.2. We are asked to find the probability that he misses for the first time on his fifth attempt. This scenario follows a geometric distribution, where the probability mass function is given by \( P(X = k) = (1-p)^{k-1} \cdot p \), where \( p \) is the probability of success. For a miss (success is considered getting a miss): \[ P(X = 5) = (0.8)^4 \cdot 0.2 \approx 0.08192 \]
03

Use Geometric Distribution for Part (b)

We need to determine the probability that the player makes his first basket on his fourth shot. Here, success means making a basket, with a probability of 0.8. The geometric distribution is applied: \[ P(X = 4) = (0.2)^3 \cdot 0.8 \approx 0.0064 \]
04

Determine Probability for Part (c) Using Complementary Counting

We must find the probability that the player makes his first basket on one of the first three shots. It's easier to calculate the complement: the probability he does not make any baskets in the first three shots, then subtract from 1. The probability of missing all first 3 shots is:\[ (0.2)^3 = 0.008 \]Thus, the probability of making at least one basket in the first three shots is:\[ 1 - 0.008 = 0.992 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Probability is a fundamental concept in mathematics and statistics. It helps us to measure how likely an event is to occur. In the exercise, the basketball player's shot success is represented by a probability of 80% or 0.8. This means that out of every 100 shots, he is expected to make 80.

The probability of an event, such as making a basket, ranges between 0 and 1.
  • A probability of 0 indicates that the event will not happen.
  • A probability of 1 means that the event is certain to occur.
  • A 0.8 probability suggests a high likelihood of success.

To calculate probabilities in various scenarios, one must understand the underlying conditions, such as whether the events are independent or dependent, which leads us to the next two concepts.
Independent Events
Independent events are those where the occurrence of one event does not affect the probability of the other event happening. In other words, each event has its own chance of occurring, and past outcomes do not influence future ones.

In our basketball exercise, the shots are considered independent. What this means is that making or missing a shot doesn't change the probability of success for subsequent shots.
  • Even if the player misses four shots in a row, the probability of making the fifth shot remains at 80%.
  • Similarly, each shot he takes has a 20% chance of missing, regardless of previous outcomes.

This independence assumption simplifies calculations and is a crucial consideration when applying geometric distribution in this context.
Complementary Counting
Complementary counting is a strategic approach used to simplify probability calculations. Instead of directly finding the probability of an event occurring, we calculate the probability of its complement (the event not happening) and subtract from 1.

In the basketball problem, complementary counting was used in part (c). The task was to find the likelihood of the player making at least one basket in his first three shots. Calculating multiple probabilities directly can be complex, so complementary counting provides a simpler way.
  • First, determine the chance that he misses all three shots:
    (0.2)*(0.2)*(0.2) = 0.008 (this is the complement).
  • Then, subtract this from 1 to find the probability of at least one success:
    1 - 0.008 = 0.992.

This method is highly useful when dealing with complex or multi-step problems where calculating all possibilities directly becomes cumbersome.

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