Problem 1E In a binomial experiment, what does it mean to say that each trial is independent of the other trials?
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Textbook Solutions for Elementary Statistics: Picturing the World
Question
Multinomial Experiments In Exercises 33 and 34, use the information below. A multinomial experiment is a probability experiment that satisfies these conditions. 1. The experiment has a fixed number of trials n, where each trial is independent of the other trials. 2. Each trial has k possible mutually exclusive outcomes: E1, E2, E3, . . ., Ek . 3. Each outcome has a fixed probability. So, P1E12 = p1, P1E22 = p2, P1E32 = p3, . . ., P1Ek2 = pk. The sum of the probabilities for all outcomes is p1 + p2 + p3 + g+ pk = 1. 4. The number of times E1 occurs is x1, the number of times E2 occurs is x2, the number of times E3 occurs is x3, and so on. 5. The discrete random variable x counts the number of times x1, x2, x3, . . ., xk occur in n independent trials where x1 + x2 + x3 + g+ xk = n. The probability that x will occur is P1x2 = n! x1!x2!x3! gxk! p1 x1 p2 x2 p3 x3gpk xk.Genetics According to a theory in genetics, when tall and colorful plantsare crossed with short and colorless plants, four types of plants will result: talland colorful, tall and colorless, short and colorful, and short and colorless,with corresponding probabilities of 916, 316, 316, and 116. Ten plants are selected.Find the probability that 5 will be tall and colorful, 2 will be tall and colorless,2 will be short and colorful, and 1 will be short and colorless.
Solution
The first step in solving 4.2 problem number 33 trying to solve the problem we have to refer to the textbook question: Multinomial Experiments In Exercises 33 and 34, use the information below. A multinomial experiment is a probability experiment that satisfies these conditions. 1. The experiment has a fixed number of trials n, where each trial is independent of the other trials. 2. Each trial has k possible mutually exclusive outcomes: E1, E2, E3, . . ., Ek . 3. Each outcome has a fixed probability. So, P1E12 = p1, P1E22 = p2, P1E32 = p3, . . ., P1Ek2 = pk. The sum of the probabilities for all outcomes is p1 + p2 + p3 + g+ pk = 1. 4. The number of times E1 occurs is x1, the number of times E2 occurs is x2, the number of times E3 occurs is x3, and so on. 5. The discrete random variable x counts the number of times x1, x2, x3, . . ., xk occur in n independent trials where x1 + x2 + x3 + g+ xk = n. The probability that x will occur is P1x2 = n! x1!x2!x3! gxk! p1 x1 p2 x2 p3 x3gpk xk.Genetics According to a theory in genetics, when tall and colorful plantsare crossed with short and colorless plants, four types of plants will result: talland colorful, tall and colorless, short and colorful, and short and colorless,with corresponding probabilities of 916, 316, 316, and 116. Ten plants are selected.Find the probability that 5 will be tall and colorful, 2 will be tall and colorless,2 will be short and colorful, and 1 will be short and colorless.
From the textbook chapter Binomial Distributions you will find a few key concepts needed to solve this.
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