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Travka [436]
3 years ago
15

If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many

M&M’s will the students need to select? (Round your answer to two decimal places.)
Mathematics
1 answer:
kotykmax [81]3 years ago
4 0

Question:

According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select peanut M&M's from an extra-large bag looking for a red candy.

If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.).

Answer:

If every student in the large Statistics class selects peanut M&M’s at random until they get a red candy the expected value of the number of  M&M’s the students need to select is 8.33 M&M's.

Step-by-step explanation:

To solve the question, we note that the statistical data presents a geometric mean. That is the probability of success is of the form.

The amount of repeated Bernoulli trials required before n eventual success outcome or

The probability of having a given number of failures before the first success is recorded.

In geometric distribution, the probability of having an eventual successful outcome depends on the the completion of a certain number of attempts with each having the same probability of success.

If the probability of each of the preceding trials is p and the kth trial is the  first successful trial, then the probability of having k is given by

Pr(X=k) = (1-p)^{k-1}p  

The number of expected independent trials to arrive at the first success for a variable Xis 1/p where p is the expected success of each trial hence p is the probability for the red and the expected value of the number of trials is 1/p or where p = 12 % which is 0.12

1/p = 1/0.12 or 25/3 or 8.33.

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3 0
3 years ago
How do you do this. Please give a full answer and explain why.
OverLord2011 [107]

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u>

\displaystyle \large{ {a}^{ - n}  =  \frac{1}{ {a}^{n} } }

Compare the terms.

\displaystyle \large{ {a}^{ - n}  =   {( - 2)}^{ - 3} }

Therefore, a = -2 and n = 3. From the law of exponent above, we receive:

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{ {( - 2)}^{ 3} } }

<u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>D</u><u>e</u><u>f</u><u>.</u> (For cubic)

\displaystyle \large{ {a}^{3}  = a \times a \times a }

Factor (-2)^3 out.

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{( - 2) \times ( - 2) \times ( - 2)}}

(-2) • (-2) = 4 | Negative × Negative = Positive.

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{4 \times ( - 2)}}

4 • (-2) = -8 | Negative Multiply Positive = Negative.

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{  - 8}}

If either denominator or numerator is in negative, it is the best to write in the middle or between numerator and denominators.

Hence,

\displaystyle \large \boxed{ {( - 2)}^{ - 3}  =  -  \frac{1}{  8}}

The answer is - 1 / 8

3 0
3 years ago
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