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Compute the following binomial probabilities directly from the formula for \(b(x ; n, p)\) : a. \(b(3 ; 8,-35)\) b. \(b(5 ; 8, .6)\) c. \(P(3 \leq X \leq 5)\) when \(n=7\) and \(p=.6\) d. \(P(1 \leq X)\) when \(n=9\) and \(p=.1\)

Short Answer

Expert verified
a. Invalid probability b. 0.187 c. 0.7571 d. 0.6126

Step by step solution

01

Understand the Binomial Probability Formula

The binomial probability formula is given by \( b(x ; n, p) = \binom{n}{x} p^x (1-p)^{n-x} \), where \(n\) is the number of trials, \(x\) is the number of successes, and \(p\) is the probability of success on each trial. \(\binom{n}{x}\) is the binomial coefficient, calculated as \(\frac{n!}{x!(n-x)!}\).
02

Calculate Probability for Part a

For part \(a\), compute \(b(3; 8, -35)\). First, calculate \(\binom{8}{3} = 56\). Now substitute into the formula: \(b(3; 8, -35) = 56 \times (-35)^3 \times (1 - (-35))^{5}\). The probabilities calculated here involve a negative value, which is invalid for \(p\). Therefore, part \(a\) is not applicable as the probability \(p\) must be between 0 and 1.
03

Calculate Probability for Part b

For part \(b\), compute \(b(5; 8, 0.6)\). Calculate \(\binom{8}{5} = 56\). Substitute into the formula: \(b(5; 8, 0.6) = 56 \times 0.6^5 \times 0.4^3\). Evaluate this to find \(b(5; 8, 0.6) = 56 \times 0.07776 \times 0.064 = 0.187\).
04

Solve for Part c (Range Probability)

To find \(P(3 \leq X \leq 5)\) for \(n=7\) and \(p=0.6\), compute \(b(3; 7, 0.6)\), \(b(4; 7, 0.6)\), and \(b(5; 7, 0.6)\).- \(b(3; 7, 0.6) = \binom{7}{3} \times 0.6^3 \times 0.4^4 = 35 \times 0.216 \times 0.0256 = 0.1935\).- \(b(4; 7, 0.6) = \binom{7}{4} \times 0.6^4 \times 0.4^3 = 35 \times 0.1296 \times 0.064 = 0.2916\).- \(b(5; 7, 0.6) = \binom{7}{5} \times 0.6^5 \times 0.4^2 = 21 \times 0.07776 \times 0.16 = 0.272\).Add the probabilities: \(P(3 \leq X \leq 5) = 0.1935 + 0.2916 + 0.272 = 0.7571\).
05

Determine Probability for Part d

For \(P(1 \leq X)\) with \(n=9\) and \(p=0.1\), calculate \(1 - P(X = 0)\) because \(P(1 \leq X) = 1 - P(X = 0)\). Compute \(b(0; 9, 0.1) = \binom{9}{0} \times 0.1^0 \times 0.9^9 = 1 \times 1 \times 0.387420489 = 0.3874\). Thus, \(P(1 \leq X) = 1 - 0.3874 = 0.6126\).

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Key Concepts

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

Binomial coefficient
The binomial coefficient is a critical concept in understanding binomial probability. It represents the number of ways to choose successes from trials and is denoted as \(\binom{n}{x}\). This can be calculated using the formula \(\frac{n!}{x!(n-x)!}\), where \(n!\) (read as \(n\) factorial) is the product of all positive integers up to \(n\).
For example, in part (b) of the original exercise, we calculated the binomial coefficient for \(\binom{8}{5}\). Here, \(\binom{8}{5} = \frac{8!}{5!3!} = 56\). This tells us there are 56 ways to achieve 5 successes in 8 trials.
Understanding the binomial coefficient helps break down the binomial probability formula into easier chunks to manage and solve.
Range probability
Range probability involves calculating the probability of outcomes falling within a certain range of interest, instead of a single value. In the original exercise, part (c) asked us to compute the probability of having between 3 and 5 successes.
To solve this, we calculate the individual probabilities for each possible outcome— 3, 4, and 5 in this case— and then sum these probabilities.
  • Calculate \(b(3; 7, 0.6)\) = 0.1935
  • Calculate \(b(4; 7, 0.6)\) = 0.2916
  • Calculate \(b(5; 7, 0.6)\) = 0.272
Add these probabilities to find the range probability: \(P(3 \leq X \leq 5) = 0.7571\).
By dividing a complex problem into smaller parts and analyzing each part separately, range probability becomes easier to understand and execute.
Probability formula
The probability formula for a binomial distribution is given by \(b(x; n, p) = \binom{n}{x} p^x (1-p)^{n-x}\). This formula helps determine the likelihood of achieving \(x\) successes in \(n\) independent trials where \(p\) is the probability of success in each trial.
The formula combines several elements: the binomial coefficient \(\binom{n}{x}\), the probability of the successes \(p^x\), and the probability of the failures \((1-p)^{n-x}\).
Let's take part (b) from the exercise as an illustration. We computed \(b(5; 8, 0.6)\) using this formula by first finding the binomial coefficient \(\binom{8}{5} = 56\), then \(0.6^5 = 0.07776\), and finally \(0.4^3 = 0.064\). Multiplying these values gives the probability: 0.187. By substituting values into the formula, you methodically work through each component to solve for the probability.
Invalid probability
In any probability question, it's critical to recognize when a probability condition is invalid. A probability \(p\) must lie between 0 and 1, inclusive, because it represents a chance of occurrence which cannot logically exceed certainty (1) or fall below impossibility (0).
In part (a) of the exercise, the given probability of \(-35\) is invalid because it doesn't fit within this accepted range. A negative probability term is not applicable in real-world scenarios. Understanding the constraints of probability is vital to correctly interpreting and solving probability problems.
Recognizing invalid probabilities helps avoid incorrect calculations and ensures the solutions are logically consistent with basic probability principles.

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Most popular questions from this chapter

The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes" (J. of Pipeline Systems Engr. and Practice, May 2012: 36-46) proposed using the Poisson distribution to model the number of failures in pipelines of various types. Suppose that for cast-iron pipe of a particular length, the expected number of failures is 1 (very close to one of the cases considered in the article). Then \(X\), the number of failures, has a Poisson distribution with \(\mu=1\). a. Obtain \(P(X \leq 5)\) by using Appendix Table A.2. b. Determine \(P(X=2)\) first from the pmf formula and then from Appendix Table A.2. c. Determine \(P(2 \leq X \leq 4)\). d. What is the probability that \(X\) exceeds its mean value by more than one standard deviation?

A new battery's voltage may be acceptable \((A)\) or unacceptable \((U)\). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(90 \%\) of all batteries have acceptable voltages. Let \(Y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is, \(P(Y=2)\) ? b. What is \(p(3)\) ? [Hint: There are two different outcomes that result in \(Y=3\).] c. To have \(Y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(Y=5\) and then determine \(p(5)\). d. Use the pattern in your answers for parts (a)-(c) to obtain a general formula for \(p(y)\).

A second-stage smog alert has been called in a certain area of Los Angeles County in which there are 50 industrial firms. An inspector will visit 10 randomly selected firms to check for violations of regulations. a. If 15 of the firms are actually violating at least one regulation, what is the pmf of the number of firms visited by the inspector that are in violation of at least one regulation? b. If there are 500 firms in the area, of which 150 are in violation, approximate the pmf of part (a) by a simpler pmf. c. For \(X=\) the number among the 10 visited that are in violation, compute \(E(X)\) and \(V(X)\) both for the exact pmf and the approximating pmf in part (b).

The \(n\) candidates for a job have been ranked \(1,2,3, \ldots, n\). Let \(X=\) the rank of a randomly selected candidate, so that \(X\) has pmf $$ p(x)= \begin{cases}1 / n & x=1,2,3, \ldots, n \\ 0 & \text { otherwise }\end{cases} $$ (this is called the discrete uniform distribution). Compute \(E(X)\) and \(V(X)\) using the shortcut formula. [Hint: The sum of the first \(n\) positive integers is \(n(n+1) / 2\), whereas the sum of their squares is \(n(n+1)(2 n+1) / 6 .]\)

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose this number \(X\) has a Poisson distribution with parameter \(\mu=.2\). (Suggested in "Average Sample Number for Semi-Curtailed Sampling Using the Poisson Distribution," \(J\). Quality Technology, 1983: 126-129.) a. What is the probability that a disk has exactly one missing pulse? b. What is the probability that a disk has at least two missing pulses? c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

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