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3241004551 [841]
3 years ago
13

A jar of candy has 6 cinnamon, 5 peppermint and 7 spearmint candies in it. Your pick five pieces of candy out of the jar at the

same time. What is the probability that three are cinnamon and two are peppermint?
Mathematics
1 answer:
leva [86]3 years ago
4 0

Answer:

2.33% probability that three are cinnamon and two are peppermint

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the candies are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

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)!}

Desired outcomes:

3 cinnamon, from a set of 6.

2 peppermint, from a set of 5. So

D = C_{6,3}*C_{5,2} = \frac{6!}{3!(6-3)!}*\frac{5!}{2!(5-2)!} = 200

Total outcomes:

5 candies, from a set of 6+5+7 = 18. So

T = C_{18,5} = \frac{18!}{5!(18-5)!} = 8568

Probability:

p = \frac{D}{T} = \frac{200}{8568} = 0.0233

2.33% probability that three are cinnamon and two are peppermint

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Answer:

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Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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