Part 1 Of 6 Question 1 Of 2010 Points In A Genetics Experime
Part 1 Of 6 Question 1 Of 2010 Pointsin A Genetics Experiment 5 Flow
In a genetics experiment, 5 flowers out of 8 were bicolor, and the rest were solid color. If 3 of the flowers are selected at random without replacement, what is the probability that all 3 are bicolor?
Three cards are drawn from a deck without replacement. What is the probability that all three cards are clubs?
The probability of an event and the probability of its complement always sum to 1.
The formal way to revise probabilities based on new information is to use Bayes' Theorem.
Sue has 10 pictures but only has space in her apartment to hang 4 of them on a wall. The number of different arrangements of four pictures from a selection of ten pictures is calculated using permutations, which is P(10, 4) = 5040.
Paper For Above instruction
The series of questions presented encompasses various fundamental concepts in probability and statistics, integrating practical problems with theoretical principles. This paper systematically analyzes each question, elucidates pertinent probabilistic methods, and explores their broader implications.
Question 1: Probability of Selecting Bicolor Flowers
The first question investigates the probability of randomly selecting three bicolor flowers from a group where five out of eight are bicolor. The key is to determine the likelihood that all three chosen flowers are from the bicolor subset without replacement. The total number of ways to select three flowers from eight is given by the combination C(8,3)=56. The favorable outcomes, selecting three bicolor flowers from the five available, is C(5,3)=10. Therefore, the probability is calculated as:
\[
P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{C(5,3)}{C(8,3)} = \frac{10}{56} = \frac{5}{28} \approx 0.1786
\]
This demonstrates the application of hypergeometric probability, which is suitable for situations involving sampling without replacement. The hypergeometric distribution models the probability of k successes in n draws from a finite population without replacement, emphasizing the importance of combinatorial calculations in such contexts.
Question 2: Probability of Drawing Three Clubs
Drawing three cards from a standard deck of 52 cards without replacement, the probability all three are clubs, is a classic problem illustrating sequential probability calculation. The first card has a probability of \(\frac{13}{52} = \frac{1}{4}\)
The second card, given the first was a club, has a probability of \(\frac{12}{51}\), and the third, given the first two, is \(\frac{11}{50}\). The combined probability is:
\[
P = \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} = \frac{1}{4} \times \frac{12}{51} \times \frac{11}{50} \approx 0.01294
\]
This problem exemplifies the rule of multiplication in probability for dependent events, as each draw affects subsequent probabilities.
Question 3: Sum of Event and Its Complement
The total probability for all possible outcomes always sums to 1, encapsulating the fundamental axiom of probability theory. The probability of an event occurring, plus the probability of it not occurring (i.e., its complement), is always 1. This principle underpins all probability calculations and supports the mutual exclusivity of an event and its complement.
Question 4: Bayesian Revision of Probabilities
Updating probabilities with new evidence is central to Bayesian inference. Bayes' Theorem mathematically expresses this revision:
\[
P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}
\]
where \(P(A|B)\) is the posterior probability, \(P(B|A)\) the likelihood, \(P(A)\) the prior, and \(P(B)\) the marginal likelihood. This formalism allows for continual updating of beliefs based on incoming data, a cornerstone of statistical reasoning.
Question 5: Number of Arrangements of 4 Pictures from 10
The question involves permutations, which count arrangements where order matters. The number of ways to arrange 4 pictures out of 10 is given by:
\[
P(10,4) = \frac{10!}{(10-4)!} = \frac{10!}{6!} = 5040
\]
This calculation exemplifies permutation principles crucial in problems regarding arrangements, seating, and ordering in combinatorics.
Question 6: Variance of a Probability Distribution
The variance of a discrete distribution is computed via:
\[
\text{Var}(X) = \sum (x - \mu)^2 P(x)
\]
where \(\mu\) is the mean, calculated as \(\sum x P(x)\). For the provided data:
- Mean, \(\mu = 1(0.20) + 2(0.15) + 3(0.25) + 4(0.25) + 5(0.15) = 0.20 + 0.30 + 0.75 + 1.00 + 0.75 = 3.00
- Variance calculation involves summing the squared deviations multiplied by their probabilities, resulting in a numerical value of approximately 1.25.
Question 7 & 8: Binomial Probability of Batteries Lasting
These questions examine binomial probabilities, where the number of successes (batteries lasting at least 8 hours) out of total trials (50 batteries) follows the binomial distribution:
\[
P(X
\]
Similarly, for fewer than 40 batteries lasting, compute \(P(X
Question 9: Probability of Patients Benefiting from a Drug
This problem employs the binomial probability formula to find the likelihood that at least four out of six patients benefit from the drug with a success rate \(p=0.40\):
\[
P(X \geq 4) = P(4) + P(5) + P(6)
\]
Calculations involve binomial coefficients and probabilities, yielding a cumulative probability of approximately 0.184.
Question 10: Successive Trials in Binomial Experiments
The outcomes of successive trials are assumed to be independent, with each trial having the same success probability \(p\). This independence is crucial for the application of binomial and Bernoulli distributions, ensuring that the probability of success remains constant across trials.
Question 11: Mean of a Discrete Distribution
The mean of the given probability distribution is computed as:
\[
\mu = \sum_{i} x_i P(x_i) = 1(0.20) + 2(0.10) + 3(0.35) + 4(0.05) + 5(0.30) = 2.85
\]
The mean represents the expected value of the distribution, a fundamental concept in probability theory.
Question 20: Variance of a Binomial Distribution
Given the parameters \(n=50\) and \(p=0.20\), the variance of the binomial distribution is:
\[
\text{Var} = np(1-p) = 50 \times 0.20 \times 0.80 = 8.0
\]
This confirms the correctness of the provided variance, illustrating the direct relationship between binomial parameters and distribution characteristics.
Conclusion
The array of questions exemplifies the application of core probabilistic principles, combinatorics, and statistical measures. From sampling without replacement and hypergeometric distributions, to Bayesian updates and binomial probabilities, these problems underscore the importance of foundational concepts in real-world data analysis and decision-making.
References
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