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Chapter 5: Problem 12

Let \(x\) be a binomial random variable with \(n=7\) and \(p=.5 .\) Find the values of the quantities in Exercises \(11-15 .\) $$ P(x \leq 1) $$

Short Answer

Expert verified
Answer: The probability is approximately 0.0625.

Step by step solution

01

Recall the binomial probability formula

The binomial probability formula is given by: $$ P(X=k) = \binom{n}{k} p^k (1-p)^{(n-k)} $$ Where \(n\) is the total number of trials, \(k\) is the number of successes, and \(p\) is the probability of success.
02

Calculate the probability for x = 0

We will use the binomial probability formula to calculate the probability of having 0 successes in 7 trials. In this case, n = 7, k = 0, and p = 0.5: $$ P(X=0) = \binom{7}{0} (0.5)^0 (1-0.5)^{(7-0)} = \frac{7!}{0!(7-0)!}(0.5)^7 = (1)(1)(0.5)^7 \approx 0.0078 $$
03

Calculate the probability for x = 1

Now, we will calculate the probability of having 1 success in 7 trials. In this case, n = 7, k = 1, and p = 0.5: $$ P(X=1) = \binom{7}{1} (0.5)^1 (1-0.5)^{(7-1)} = \frac{7!}{1!(7-1)!}(0.5)^7 = (7)(0.5)(0.5)^6 \approx 0.0547 $$
04

Calculate the probability \(P(x \leq 1)\)

Finally, to find the probability \(P(x \leq 1)\), we will add the probabilities calculated in Steps 2 and 3: $$ P(x \leq 1) = P(X=0) + P(X=1) \approx 0.0078 + 0.0547 \approx 0.0625 $$ So, the probability of having at most 1 success in 7 trials is approximately 0.0625.

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Key Concepts

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

Binomial Random Variable
In probability theory and statistics, a binomial random variable is a specific type of discrete random variable. It counts how often a particular event, usually referred to as a 'success,' occurs in a fixed number of trials or experiments.

Each trial is independent, meaning the outcome of one trial does not affect another, and there are only two possible outcomes: success or failure. The two main parameters that define a binomial random variable are the number of trials, denoted by the letter 'n', and the probability of success in each trial, denoted by the letter 'p'. For instance, in a coin toss, if we define 'success' as landing heads, and we flip the coin 7 times, this scenario can be described by a binomial random variable.

The formula to find the probability of obtaining exactly 'k' successes in 'n' trials is given by the binomial probability formula. As applied to our coin example, to find the probability of achieving exactly one head (success) in seven tosses (trials), we would use this formula to perform the calculations.
Binomial Distribution
A binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent values under a given set of parameters or assumptions. Essentially, it reflects the distribution of a binomial random variable.

Visualizing this distribution, we often use a histogram or a probability mass function plot, where each bar represents the probability of obtaining each possible number of successes (from 0 up to 'n'). The shape of the distribution is determined by the parameters 'n' and 'p'. As 'n' increases or as 'p' gets closer to 0.5, the distribution tends to look more symmetrical.

For practical applications, knowing the binomial distribution helps in understanding the variability and possible outcomes of processes that follow a binomial pattern - for example, quality control testing in manufacturing, where 'success' could be a product passing inspection.
Probability of Success
In the context of binomial variables and distribution, the probability of success 'p' is a crucial concept. It is the likelihood that any given trial will result in a success. This probability is assumed to be constant for each trial within the series.

If we return to the coin example, assuming a fair coin, the probability of getting heads (our defined 'success') is 0.5, because there is an equal chance of landing heads or tails. The 'probability of success' directly affects the possible outcomes, and it is essential to accurately determine this probability when analyzing real-world situations.

To calculate cumulative probabilities, like 'What is the probability of achieving one or fewer heads in seven coin tosses?', you sum the individual probabilities of all events from 0 up to your desired number of successes, computing each by employing the binomial formula with our known 'n' and 'p'. Depending on the values of 'n' and 'p', these cumulative probabilities can offer significant insights into the likelihood of a range of outcomes and are highly informative for predicting and understanding the behavior of binomial processes.

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Most popular questions from this chapter

Draw three cards randomly from a standard deck of 52 cards and let \(x\) be the number of kings in the draw. Evaluate the probabilities and answer the questions in Exercises \(26-28\) \(p(x)\) for \(x=0,1,2,3\)

For the random variables described, find and graph the probability distribution for \(x .\) Then calculate the mean, variance, and standard deviation. Toss a pair of dice and record \(x,\) the sum of the numbers on the two upper faces.

Let \(x\) be a binomial random variable with \(n=20\) and \(p=.1\). a. Calculate \(P(x \leq 4)\) using the binomial formula. b. Calculate \(P(x \leq 4)\) using Table 1 in Appendix I. c. Use the following Excel output to calculate \(P(x \leq 4)\). Compare the results of parts a, b, and c. d. Calculate the mean and standard deviation of the random variable \(x\). e. Use the results of part d to calculate the intervals \(\mu \pm \sigma, \mu \pm 2 \sigma,\) and \(\mu \pm 3 \sigma .\) Find the probability that an observation will fall into each of these intervals. f. Are the results of part e consistent with Tchebysheff's Theorem? With the Empirical Rule? Why or why not? Excel output for Exercise 33: Binomial with \(n=20\) and \(p=.1\) $$ \begin{array}{|c|c|c|c|c|} \hline & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} \\ \hline 1 & \mathrm{x} & \mathrm{p}(\mathrm{x}) & \mathrm{x} & \mathrm{p}(\mathrm{x}) \\ \hline 2 & 0 & 0.1216 & 11 & 7 \mathrm{E}-07 \\ \hline 3 & 1 & 0.2702 & 12 & 5 \mathrm{E}-08 \\ \hline 4 & 2 & 0.2852 & 13 & 4 \mathrm{E}-09 \\ \hline 5 & 3 & 0.1901 & 14 & 2 \mathrm{E}-10 \\ \hline 6 & 4 & 0.0898 & 15 & 9 \mathrm{E}-12 \\ \hline 7 & 5 & 0.0319 & 16 & 3 \mathrm{E}-13 \\ \hline 8 & 6 & 0.0089 & 17 & 8 \mathrm{E}-15 \\ \hline 9 & 7 & 0.0020 & 18 & 2 \mathrm{E}-16 \\ \hline 10 & 8 & 0.0004 & 19 & 2 \mathrm{E}-18 \\ \hline 11 & 9 & 0.0001 & 20 & 1 \mathrm{E}-20 \\ \hline 12 & 10 & 0.0000 & & \\ \hline \end{array} $$

A key ring contains four office keys that are identical in appearance, but only one will open your office door. Suppose you randomly select one key and try it. If it does not fit, you randomly select one of the three remaining keys. If that key does not fit, you randomly select one of the last two. Each different sequence that could occur in selecting the keys represents a set of equally likely simple events. a. List the simple events in \(S\) and assign probabilities to the simple events. b. Let \(x\) equal the number of keys that you try before you find the one that opens the door \((x=1,2,3,4)\). Then assign the appropriate value of \(x\) to each simple event. c. Calculate the values of \(p(x)\) and display them in a table. d. Construct a probability histogram for \(p(x)\).

Let \(x\) be the number of successes observed in a sample of \(n=5\) items selected from a population of \(N=10 .\) Suppose that of the \(N=10\) items, \(M=6\) are considered "successes." Find the probabilities in Exercises \(11-13 .\) The probability of observing exactly two successes.
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