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Evaluate these binomial probabilities: a. \(C_{2}^{8}(.3)^{2}(.7)^{6}\) b. \(C_{0}^{4}(.05)^{0}(.95)^{4}\) c. \(C_{3}^{10}(.5)^{3}(.5)^{7}\) d. \(C_{1}^{7}(.2)^{1}(.8)^{6}\)

Short Answer

Expert verified
In summary, the calculated binomial probabilities are as follows: a. \(P(X=2) \approx 0.2965\) b. \(P(X=0) \approx 0.8145\) c. \(P(X=3) \approx 0.1172\) d. \(P(X=1) \approx 0.4398\)

Step by step solution

01

a. The first binomial probability calculation

Recall the formula for binomial probability: \(P(X=k) = C_{k}^{n}p^k(1-p)^{n-k}\). Here, we are given \(n = 8\), \(k = 2\), and the probability \(p = 0.3\). Therefore, we will calculate the first probability as follows: \(P(X=2) = C_{2}^{8}(.3)^{2}(.7)^{6} = \frac{8!}{2!6!}(0.3)^{2}(0.7)^{6}\). Now we calculate the factorial values and plug them in: \(P(X=2) = \frac{40320}{2*720}(0.09)(0.7)^{6} = \frac{40320}{1440}(0.09)(0.7)^{6}\). Next, simplify the fraction and compute the probability: \(P(X=2) = 28(0.09)(0.7)^{6} \approx 0.2965.\) So the answer is \(P(X=2) \approx 0.2965\). #b. \(C_{0}^{4}(.05)^{0}(.95)^{4}\)#
02

b. The second binomial probability calculation

This time, we have \(n = 4\), \(k = 0\), and \(p = 0.05\). We use the binomial probability formula to calculate the probability: \(P(X=0) = C_{0}^{4}(.05)^{0}(.95)^{4} = \frac{4!}{0!4!}(1)(0.95)^{4}\). Calculate the factorial values, plug them in, and simplify: \(P(X=0) = \frac{24}{1*24}(1)(0.95)^{4} = 1(0.95)^{4} \approx 0.8145.\) So the answer is \(P(X=0) \approx 0.8145\). #c. \(C_{3}^{10}(.5)^{3}(.5)^{7}\)#
03

c. The third binomial probability calculation

Here, we have \(n = 10\), \(k = 3\), and \(p = 0.5\). As before, use the binomial probability formula to calculate: \(P(X=3) = C_{3}^{10}(.5)^{3}(.5)^{7} = \frac{10!}{3!7!}(0.5)^{3}(0.5)^{7}\). Calculate the factorial values, plug them in, and simplify: \(P(X=3) = \frac{3628800}{6*5040}(0.5)^{3}(0.5)^{7} = \frac{3628800}{30240}(0.5)^{10}\). Next, compute the probability: \(P(X=3) = 120(0.5)^{10} \approx 0.1172\). So the answer is \(P(X=3) \approx 0.1172\). #d. \(C_{1}^{7}(.2)^{1}(.8)^{6}\)#
04

d. The fourth binomial probability calculation

For this problem, we have \(n = 7\), \(k = 1\), and \(p = 0.2\). Again, use the binomial probability formula: \(P(X=1) = C_{1}^{7}(.2)^{1}(.8)^{6} = \frac{7!}{1!6!}(0.2)^{1}(0.8)^{6}\). Calculate the factorial values, plug them in, and simplify the expression: \(P(X=1) = \frac{5040}{720}(0.2)(0.8)^{6} = \frac{5040}{720}(0.2)(0.8)^{6}\). Compute the probability: \(P(X=1) = 7(0.2)(0.8)^{6} \approx 0.4398\). So the answer is \(P(X=1) \approx 0.4398\).

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Key Concepts

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

Probability Theorem
The Probability Theorem forms the backbone of many statistical calculations, especially in scenarios involving multiple outcomes. The binomial probability formula is one such application of this theorem, often employed to find the probability of a certain number of successes in a series of trials. The binomial distribution is heavily dependent on three parameters: the total number of trials (), the number of successful outcomes we are interested in (k), and the probability of success in a single trial (p).
  • Understanding this setup is crucial for applying the probability theorem accurately.
  • Each trial is independent, meaning the outcome of one does not affect the others.
The formula for finding the probability of exactly k successes in n trials is expressed as:\[ P(X = k) = C_{k}^{n} p^k (1-p)^{n-k} \]Where \( C_{k}^{n} \), often called "n choose k", represents the number of combinations, showing how many different ways k successes can occur among n trials. This is where our understanding of factorials becomes essential.
Factorials in Probability
Factorials play a significant role in calculating combinations, which is an essential part of the binomial probability formula. A factorial, denoted \( n! \), is the product of all positive integers up to n. For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). Factorials allow us to compute combinations \( C_{k}^{n} \) using the formula:\[ C_{k}^{n} = \frac{n!}{k!(n-k)!} \]The combination formula determines the number of ways to choose k successes from a pool of n trials. Here’s how it breaks down:
  • \( n! \) is the factorial of the total number of trials.
  • \( k! \) is the factorial of the number of successes.
  • \( (n-k)! \) is the factorial of the number of failures.
This factorial-based method enables accurate calculation of probabilities when combined with single trial probabilities, which are influenced by the probability theorem discussed earlier.
Probability Calculation
Calculating probabilities using the binomial probability formula involves several clear steps.
_Understanding each step is essential for accuracy:_
  • Begin by identifying n, k, and p based on the problem statement. For example, in the problem \( P(X=2) = C_{2}^{8}(.3)^{2}(.7)^{6} \), we know \( n=8 \), \( k=2 \), and \( p=0.3 \).
  • Calculate the combination \( C_{k}^{n} \), which can be found using factorials, as explained in the section above.
  • Multiply the result by \( p^k \), where \( p \) is the probability of success, raised to the number of successful trials.
  • Additionally, multiply it by \((1-p)^{n-k} \), representing the probability of failure raised to the number of failed trials.
Carrying out these steps carefully, combining them as per the formula, gives us the probability of exactly k successes. For example, simplifying \( \frac{8!}{2!6!}(0.3)^{2}(0.7)^{6} \) accurately leads to the probability of \( \approx 0.2965 \), ensuring a clear understanding of each component and operation in the calculation process.

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Most popular questions from this chapter

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