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Chapter 4: Problem 1

According to a CBS/New York Times poll taken in \(1992,15 \%\) of the public have responded to a telephone call-in poll. In a random group of five people, what is the probability that exactly two have responded to a call-in poll? (A) \(10(0.15)^{2}(0.85)^{3}\) (B) \(5(0.15)^{2}(0.85)^{3}\) (C) \((0.15)^{2}(0.85)^{3}\) (D) \((0.15)^{2}\) (E) \(5(0.15)^{2}\)

Short Answer

Expert verified
(A) 10(0.15)^2(0.85)^3

Step by step solution

01

Identify the given data

The probability that a person has responded to a call-in poll is given as 15%, which is 0.15 in decimal form. We need to find the probability that exactly two out of five randomly selected people have responded to a call-in poll.
02

Identify the binomial probability formula

The probability of exactly k successes in n trials in a binomial distribution is given by the formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where:- \( P(X = k) \) is the probability of k successes out of n trials - \( \binom{n}{k} \) is the binomial coefficient- p is the probability of success on a single trial- \( (1-p) \) is the probability of failure on a single trial
03

Compute the binomial coefficient

The binomial coefficient for choosing 2 out of 5 is: \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = 10 \]
04

Substitute values into the formula

Using n = 5, k = 2, and p = 0.15, substitute the values into the binomial probability formula: \[ P(X = 2) = \binom{5}{2} (0.15)^2 (0.85)^{5-2} \]\[ P(X = 2) = 10 (0.15)^2 (0.85)^3 \]
05

Match the calculated expression with the given options

Compare the obtained expression with the provided choices. The correct option is:(A) \(10(0.15)^2(0.85)^3\).

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Key Concepts

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

probability
Probability helps us measure the chance of an event happening. In everyday language, it answers questions like 'What are the odds?' or 'How likely is it?'.
In mathematics, we express this likelihood as a number between 0 and 1.
If something is impossible, its probability is 0. If it's certain, its probability is 1.
For example, the probability of flipping a fair coin and getting heads is 0.5, because there are two equally likely outcomes: heads or tails.
Therefore, when we talk about probability, we are dealing with possible outcomes and trying to predict how frequently each one might occur.
binomial distribution
A binomial distribution describes the number of successes in a fixed number of trials.
Each trial has two possible outcomes: success or failure.
This distribution is useful for situations where you repeat an experiment multiple times, like flipping a coin.
The parameters of a binomial distribution are:
  • n: The number of trials (e.g., flipping a coin 5 times).
  • p: The probability of success in a single trial (e.g., the probability of getting heads is 0.5).
In our exercise, responding to a call-in poll is the 'success', and we have 5 trials.
binomial coefficient
The binomial coefficient \(\binom{n}{k}\) tells us the number of ways we can choose 'k' successes out of 'n' trials.
It's often read as 'n choose k'.
Mathematically, it鈥檚 determined by the formula:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
Here, '!' represents a factorial, which means multiplying a series of descending natural numbers.
For example, the factorial of 5 (written as 5!) is 5 x 4 x 3 x 2 x 1 = 120.
Using this understanding, in our exercise, we compute the binomial coefficient for choosing 2 out of 5 as 10.
probability formula
The probability formula for exactly k successes in a binomial distribution is:
\[\binom{n}{k} p^k (1-p)^{n-k}\]
Here, \[ \binom{n}{k} \] is the binomial coefficient, p is the probability of success, and \ [(1-p)] \ is the probability of failure.
We use this formula to find the probability of specific outcomes.
In our exercise, we substituted:\
  • n = 5: total trials
  • k = 2: successes
  • p = 0.15: success probability
to get:
\[ P(X = 2) = 10 (0.15)^2 (0.85)^3 \]
This shows us the structure and crucial parts of the calculation.

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Most popular questions from this chapter

Suppose a retailer knows that the mean number of broken eggs per carton is 0.3 with a standard deviation of 0.18 . In a shipment of 100 cartons, what is the expected number of broken eggs and what is the standard deviation? Assume independence between cartons. \(\begin{array}{ll}\text { (A) } E \text { (broken) }=3 & \text { SD(broken) }=1.8\end{array}\) (B) \(E(\) broken \()=30 \quad\) SD(broken) \(=1.8\) \(\begin{array}{ll}\text { (C) } E \text { (broken) }=30 & \text { SD(broken) }=18\end{array}\) (D) \(E\) (broken) \(=300 \quad\) SD(broken) \(=18\) (E) \(E(\) broken \()=300 \quad\) SD(broken \()=180\)

Suppose that one out of 20 apples from a particular orchard is wormy. What are the mean and standard deviation for the number of apples to be sampled from this orchard before finding a wormy apple? (A) Mean \(=5,\) standard deviation \(=1\) (B) Mean \(=5,\) standard deviation \(=1-0.05\) (C) Mean \(=10,\) standard deviation \(=\sqrt{\frac{(0.05)^{2}}{1-0.05}}\) (D) Mean \(=20,\) standard deviation \(=\sqrt{380}\) (E) Mean \(=20,\) standard deviation \(=\sqrt{\frac{(0.05)^{2}}{1-0.05}}\)

An inspection procedure at a manufacturing plant involves picking three items at random and then accepting the whole lot if at least two of the three items are in perfect condition. If in reality \(90 \%\) of the whole lot are perfect, what is the probability that the lot will be accepted? (A) \(3(0.1)^{2}(0.9)\) (B) \(3(0.9)^{2}(0.1)\) (C) \((0.1)^{3}+(0.1)^{2}(0.9)\) (D) \((0.1)^{3}+3(0.1)^{2}(0.9)\) (E) \((0.9)^{3}+3(0.9)^{2}(0.1)\)

Given that \(52 \%\) of the U.S. population are female and \(15 \%\) are older than age 65 , can we conclude that \((0.52)(0.15)=7.8 \%\) are women older than age 65 ? (A) Yes, by the multiplication rule. (B) Yes, by conditional probabilities. (C) Yes, by the law of large numbers. (D) No, because the events are not independent. (E) No, because the events are not mutually exclusive.

At a warehouse sale, 100 customers are invited to choose one of 100 unopened, identical boxes, each containing one item. Five boxes contain $$\$ 700$$ flat- screen television sets, 25 boxes contain $$\$ 540$$ smartphones, and the remaining boxes contain $$\$ 260$$ cameras. What should a customer be willing to pay to participate in the sale? (A) $$\$ 260$$ (B) $$\$ 35^{2}$$ (C) $$\$ 500$$ (D) $$\$ 540$$ (E) $$\$ 699$$
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