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91Ó°ÊÓ

Alysha makes \(40 \%\) of her free throws. She takes five free throws in a game. If the shots are independent of each other, the probability that she misses the first two shots but makes the other three is about (a) \(0.230\). (b) \(0.115\). (c) \(0.023\). (d) \(0.600\).

Short Answer

Expert verified
(c) 0.023.

Step by step solution

01

Identify Required Probability

Identify the probability required: Alysha misses the first two shots and makes the last three. This translates into the sequence 'MMSSS' where 'M' is a miss and 'S' is a success (make).
02

Determine Individual Probabilities

Calculate the probability of missing a shot: if the probability of making a shot is 40% or 0.4, the probability of missing is 1 - 0.4 = 0.6.
03

Calculate Compound Probability

Since the shots are independent, use the formula for the probability of a sequence of independent events. The probability of missing the first two shots and making the next three is given by \[ (0.6)^{2} \times (0.4)^{3} \].
04

Compute Exact Probability

Calculate the expression from the previous step:- \((0.6)^{2} = 0.36\)- \((0.4)^{3} = 0.064\)Now multiply these values: \[ 0.36 \times 0.064 = 0.02304 \].
05

Compare to Given Options

Compare the calculated probability with the given answer choices: (c) 0.023 is the closest match to the calculated result 0.02304.

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Key Concepts

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

Independent Events
In probability and statistics, independent events are events where the outcome of one does not affect the outcome of another. Understanding this concept is crucial when dealing with exercises like the one involving Alysha's free throws.

For events to be independent, the probability of event A occurring does not change based on whether event B occurs. Mathematically, this is expressed as \( P(A \cap B) = P(A) \times P(B) \).

This principle allows us to simplify the process of calculating probabilities for sequences of events. In Alysha's case, each free throw is an independent event because her performance on one throw does not influence her performance on subsequent throws. This independence allows us to calculate the probability of a specific sequence of outcomes by simply multiplying the probabilities of the individual events.

Independent events make calculations straightforward since we don't have to adjust for previous outcomes. It allows us to determine the likelihood of multiple outcomes happening together using simple multiplication, regardless of any patterns that may occur.
Compound Probability
Compound probability involves finding the probability of two or more events happening together. In Exercises involving sequences, like Alysha's shooting scenario, compound probability is essential to determine the likelihood of specific outcomes.

Compound events are combinations of two or more simple events. Their probabilities can be calculated by multiplying the probabilities of each event. This is straightforward if the events are independent. For example, the probability of Alysha missing her first two shots and then making the next three requires determining the compound probability of the sequence 'MMSSS'.

This calculation involves multiplying the individual probabilities:
  • The probability of missing a shot: \(0.6\)
  • The probability of making a shot: \(0.4\)
Thus, the compound probability for her missing the first two shots and making the subsequent three is \((0.6)^2 \times (0.4)^3\). This mathematical representation clearly showcases how compound probabilities are derived.

Compound probabilities help answer complex questions about multiple events where calculating each probability alone wouldn't suffice.
Binomial Probability
The concept of binomial probability is vital when determining the probability of a fixed number of successes in a series of identical trials. It simplifies the process of finding probabilities when outcomes are binary, such as Alysha's free throw attempts being either a make or a miss.

Binomial probability applies to experiments with fixed numbers of trials (\(n\)), with each trial resulting in either success (with probability \(p\)) or failure (with probability \(1-p\)). The probability of exactly \(k\) successes is detailed by the formula:
\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]
For Alysha's situation, the Bernoulli trials are her free throws, each with a probability of a successful shot: \(p = 0.4\). Using the binomial probability theorem, we can calculate the chance of different sequences of makes and misses, even though in this case we used straightforward multiplication due to independence.

Understanding binomial probability enables you to quickly identify and calculate the probability of sequences like Alysha's, especially when sequences grow more complex or larger. It also shows why binomial probability is a powerful tool in statistical calculations, useful beyond simple multiplication when more detailed inference about distributions or predictions is needed.

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Most popular questions from this chapter

The \(z\) confidence interval and test are based on the sampling distribution of the sample mean \(x^{-} \bar{x}\). The National Survey of Student Engagement (NSSE) asks college seniors to rate how much their experience at their institution has contributed to their ability to think critically and analytically. Ratings are on a scale of 1 to 7 , with 7 being the highest (best) rating. Suppose scores for all seniors in 2013 are Normal with mean \(\mu=3.3\) and standard deviation \(\sigma=0.8\). (a) You take an SRS of 100 seniors. According to the \(99.7\) part of the 68-95-99.7 rule, about what range of ratings do you expect to see in your sample? (b) You look at many SRSs of size 100. About what range of sample mean ratings \(\mathrm{x}^{-} \bar{x}\) do you expect to see?

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