Multinomial Distribution - Flashcard set - Probability

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10 questions on a multiple-choice test, each with 4 answer options, and identifying how many questions have A, B, C, or D as correct answers.

The probabilities for the distribution of correct answers are:

P(x_{A}, x_{B}, x_{C}, x_{D}) = \frac{10!}{x_{A}!x_{B}!x_{C}!x_{D}!} \left(\frac{1}{4}\right)^{10}

where

x_{A} + x_{B} + x_{C} + x_{D} = 10

and it's assumed each answer is equally likely to be correct.

A survey categorizes responses into 'Agree', 'Disagree', or 'Neutral' from a sample of 50 people.

The probability of a specific number of responses in each category is:

P(x_{agree}, x_{disagree}, x_{neutral}) = \frac{50!}{x_{agree}!x_{disagree}!x_{neutral}!} p_{agree}^{x_{agree}} p_{disagree}^{x_{disagree}} p_{neutral}^{x_{neutral}}

where

x_{agree} + x_{disagree} + x_{neutral} = 50

and

p_{agree}

,

p_{disagree}

, and

p_{neutral}

are the probabilities of each response.

Randomly assigning 12 students to 4 different project teams equally likely.

The probability of a given team distribution is:

P(x_{1}, x_{2}, x_{3}, x_{4}) = \frac{12!}{x_{1}!x_{2}!x_{3}!x_{4}!} \left(\frac{1}{4}\right)^{12}

where

x_{1} + x_{2} + x_{3} + x_{4} = 12

since the teams must have a total of 12 students.

Distributing 15 identical gifts among 3 friends such that each friend gets at least one gift.

The probability of each friend receiving a certain number of gifts is:

P(x_{1}, x_{2}, x_{3}) = \frac{15!}{x_{1}!x_{2}!x_{3}!} \left(\frac{1}{3}\right)^{15}

where

x_{1} + x_{2} + x_{3} = 15

and all gifts are identical.

Choosing 20 fruits from a box containing apples, bananas, and oranges, knowing the probability of picking each type.

The multinomial distribution for the fruit choice is:

P(x_{apples}, x_{bananas}, x_{oranges}) = \frac{20!}{x_{apples}!x_{bananas}!x_{oranges}!} p_{apples}^{x_{apples}} p_{bananas}^{x_{bananas}} p_{oranges}^{x_{oranges}}

where

x_{apples} + x_{bananas} + x_{oranges} = 20

.

Drawing 5 balls from a bag containing balls in 4 different colors, with replacement, and counting the number of each color.

The probability of drawing a certain combination of colors is:

P(x_{1}, x_{2}, x_{3}, x_{4}) = \frac{5!}{x_{1}!x_{2}!x_{3}!x_{4}!} p_{1}^{x_{1}} p_{2}^{x_{2}} p_{3}^{x_{3}} p_{4}^{x_{4}}

where

x_{1} + x_{2} + x_{3} + x_{4} = 5

and each

p_i

is the probability of drawing a ball of color

i

.

Rolling a fair six-sided dice 10 times and counting the occurrences of each outcome.

The probability of any specific outcome sequence is given by the multinomial distribution formula:

P(x_1, x_2, x_3, x_4, x_5, x_6) = \frac{10!}{x_1!x_2!x_3!x_4!x_5!x_6!} \left(\frac{1}{6}\right)^{10}

where

x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 10

.

In a quality check, a batch of 30 products could be classified as High, Medium, or Low quality.

The probability of a given quality distribution is:

P(x_{High}, x_{Medium}, x_{Low}) = \frac{30!}{x_{High}!x_{Medium}!x_{Low}!} p_{High}^{x_{High}} p_{Medium}^{x_{Medium}} p_{Low}^{x_{Low}}

where

x_{High} + x_{Medium} + x_{Low} = 30

and

p_{Quality}

represents the probability of each quality level.

5 people ordering from a menu with 4 meal options, and we want to know how many times each meal is ordered.

The probability of a specific meal distribution is:

P(x_{1}, x_{2}, x_{3}, x_{4}) = \frac{5!}{x_{1}!x_{2}!x_{3}!x_{4}!} p_{1}^{x_{1}} p_{2}^{x_{2}} p_{3}^{x_{3}} p_{4}^{x_{4}}

where

x_{1} + x_{2} + x_{3} + x_{4} = 5

and

p_i

is the probability of a person choosing the

i

-th meal option.

Testing 8 electronic components which can either pass, need rework, or fail.

For a given number of each outcome:

P(x_{pass}, x_{rework}, x_{fail}) = \frac{8!}{x_{pass}!x_{rework}!x_{fail}!} p_{pass}^{x_{pass}} p_{rework}^{x_{rework}} p_{fail}^{x_{fail}}

where

x_{pass} + x_{rework} + x_{fail} = 8

and

p_{outcome}

is the probability of each test outcome.

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