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Chapter 8: Problem 18

Use the formula \(C(n, x) p^{x} q^{n-x}\) to determine the probability of the given event. Let \(X\) be the number of successes in five independent trials in a binomial experiment in which the probability of success is \(p=\frac{2}{5}\). Find: a. \(P(X=4)\) b. \(P(2 \leq X \leq 4)\)

Short Answer

Expert verified
a. \(P(X=4) = \frac{96}{625}\) b. \(P(2 \leq X \leq 4) = \frac{345}{625}\)

Step by step solution

01

Determine q (probability of failure)#

Since p is the probability of success, then q is the probability of failure, calculated as q = 1 - p. Here, p = 2/5, so we have q = 1 - (2/5) = 3/5.
02

Calculate P(X=4) by applying the formula#

To find P(X=4), replace n=5, x=4, p=2/5, and q=3/5 in the formula \(C(n,x)p^xq^{n-x}\), and calculate the result. \[P(X=4) = C(5,4)\left(\frac{2}{5}\right)^4\left(\frac{3}{5}\right)^{5-4}\] \(P(X=4) = \frac{5!}{4!(5-4)!}\left(\frac{2}{5}\right)^4\left(\frac{3}{5}\right)^1\) \(P(X=4) = 5 \cdot \left(\frac{2}{5}\right)^4\left(\frac{3}{5}\right)\) After calculating the expression, we get: \(P(X=4) = \frac{96}{625}\)
03

Calculate P(2 鈮� X 鈮� 4) by summing individual probabilities#

Since we need to find the probability of having 2, 3, or 4 successes in the 5 trials, we're going to calculate each individual probability and sum them up. \[P(2\leq X\leq 4) = P(X=2) + P(X=3) + P(X=4)\] Using the same formula as before, but replacing x with the corresponding values: \[P(X=2) = C(5,2)\left(\frac{2}{5}\right)^2\left(\frac{3}{5}\right)^{5-2}\] \[P(X=3) = C(5,3)\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^{5-3}\] Now, sum up the individual probabilities: \(P(2 \leq X \leq 4) = P(X=2) + P(X=3) + P(X=4)\) After calculating the expressions, we get: \(P(2 \leq X \leq 4) = \frac{345}{625}\) So, the answers are: a. \(P(X=4) = \frac{96}{625}\) b. \(P(2 \leq X \leq 4) = \frac{345}{625}\)

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Key Concepts

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

Probability of Success
When we talk about probability of success, we refer to the chance that an event we desire actually happens. For example, if we toss a coin, we might define success as landing on heads. The probability of heads (our success) is 0.5, because there are two possible outcomes, and one is favorable. Similarly, in a binomial experiment like the one mentioned in the exercise, success is determined by an event happening in each trial, such as getting a correct answer on a true/false question. Here, we have a fixed probability of success (denoted as p) for each trial.
Probability Formula
The probability formula is a mathematical way to calculate the likelihood of an event. In the context of binomial distribution, the specific formula used is \(C(n, x) p^{x} q^{n-x}\), where:\
    \
  • \n\ represents the total number of trials.\
    • \
  • \x\ is the number of successful events we are trying to find the probability of.\
    • \
  • \p\ is the probability of success on a single trial.\
    • \
  • \q\ is the probability of failure in a single trial (\(q = 1 - p\)).\
    • \
\Additionally, \(C(n, x)\) is a binomial coefficient representing the number of ways we can choose x successes out of n trials. Mathematically, it is expressed as \(\frac{n!}{x!(n-x)!}\). This formula allows us to calculate the probability of getting exactly x successes in n independent trials.
Independent Trials
The term independent trials is crucial in probability theory, especially within the binomial distribution concept. Independent trials mean that the outcome of one trial does not affect the outcome of another. In other words, each event is separate, and the probability of success remains constant across trials. For instance, each flip of a coin is independent of the previous flip. Similarly, in a binomial experiment, each trial (like answering a true/false question) is assumed not to interfere with the probability of answering subsequent questions correctly. This assumption is essential for the binomial probability formula to be valid.
Binomial Distribution
A binomial distribution is a type of probability distribution that has two possible outcomes (hence 'bi'-nomial) for each trial, often named as 'success' and 'failure'. In the exercise provided, a binomial distribution is utilized to model the number of successes in a series of independent trials with the same probability of success. Key characteristics include a fixed number of trials (n), only two possible outcomes for each trial, the trials are independent, and the probability of success (p) is the same for each trial. Binomial distribution answers questions like 'What is the probability of having exactly four successes in five trials?' and can be visually represented using a histogram showing the probabilities of the different numbers of successes (0 through n). Mathematically, binomial distribution makes use of factorials and combinations to calculate probabilities, which was demonstrated in the step-by-step solution for the given exercise.

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Most popular questions from this chapter

The following table gives the scores of 30 students in a mathematics examination: $$ \begin{array}{lccccc} \hline \text { Scores } & 90-99 & 80-89 & 70-79 & 60-69 & 50-59 \\ \hline \text { Students } & 4 & 8 & 12 & 4 & 2 \\ \hline \end{array} $$ Find the mean and the standard deviation of the distribution of the given data.

A probability distribution has a mean of 50 and a standard deviation of \(1.4 .\) Use Chebychev's inequality to find the value of \(c\) that guarantees the probability is at least \(96 \%\) that an outcome of the experiment lies between \(50-c\) and \(50+c .\)

The seasonally adjusted annualized sales rate for U.S. cars and light trucks, in millions of units. for May 2003 through April 2004 are given in the following tables: $$ \begin{array}{l} 2003\\\ \begin{array}{cccccccc} \hline \mathrm{M} & \mathrm{J} & \mathrm{J} & \mathrm{A} & \mathrm{S} & \mathrm{O} & \mathrm{N} & \mathrm{D} \\ \hline 16.5 & 16.5 & 17.0 & 18.5 & 17.0 & 16.0 & 17.0 & 18.0 \\ \hline \\ 2004 & & & & & & & \\ \cline { 1 - 5 } \mathrm{J} & \mathrm{F} & \mathrm{M} & \mathrm{A} & & & & \\ \hline 16.3 & 16.5 & 16.8 & 16.5 & & & & \\ \hline \end{array} \end{array} $$ What is the average seasonally adjusted annualized sales rate for U.S. motor vehicles for the period in question? What is the standard deviation for these data?

According to data released by the Chamber of Commerce of a certain city, the weekly wages (in dollars) of female factory workers are normally distributed with a mean of 575 and a standard deviation of 50 . Find the probability that a female factory worker selected at random from the city makes a weekly wage of $$\$ 550$$ to $$\$ 650$$.

In American roulette, as described in Example 6, a player may bet on a split (two adjacent numbers). In this case, if the player bets \(\$ 1\) and either number comes up, the player wins \(\$ 17\) and gets his \(\$ 1\) back. If neither comes up, he loses his \(\$ 1\) bet. Find the expected value of the winnings on a \(\$ 1\) bet placed on a split.
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