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SashulF [63]
3 years ago
13

The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’

s typically contains 52 M&M’s.Find the probability that
c.) Six M&M’s are brown. %
d.) Twenty-five M&M’s are brown. %
e.) All of the M&M’s are brown. %
f.) Would it be unusual for a package to have only brown M&M’s? If this were to happen, what would you think is the reason?
Mathematics
1 answer:
e-lub [12.9K]3 years ago
3 0

Step-by-step explanation:

a)State the random variable.x = number of brown M&Ms in the milk chocolate box.

(b)Argue that this is a binomial analysis.

The binomial analysis includes the following characteristics: 1.Fixed number of measurements.

2. The experiments are autonomous.

3. There are just two results.

4. The expectation of victory will not shift from court to court.

This is a binomial experiment since the number of experiments is set and the measures are independent of each other. In fact, there are only two results – brown M&Ms or not.

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A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the st
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Answer:

0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.

This means that n = 15, p = \frac{1}{5} = 0.2

What is the probability that he answers at least 10 questions correctly?

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001

P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011

P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0

P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0

P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0

P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111

0.0111% probability that he answers at least 10 questions correctly

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