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defon
3 years ago
8

An ice chest contains cans of apple juice, cans of grape juice, cans of orange juice, and cans of mango juice. Suppose that you

reach into the container and randomly select three cans in succession. Find the probability of selecting three cans of juice.
Mathematics
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

4.9%

Step-by-step explanation:

This question is incomplete. However; it can be found on search engines. The complete question is  as follows :

An ice chest contains cans of six apple juice, eight cans of grape juice, four cans of orange juice, and two cans of mango juice. Suppose that you reach into the container and randomly select three cans in succession. Find the probability of selecting three cans of grape juice.

Solution :

In an ice chest there are different cans of juice. Among them

Number of cans of apple juice = 6

Number of cans of grape juice = 8

Number of cans of orange juice = 4

Number of cans of mango juice = 2

Total number of cans of juice = 6 + 8 + 4 + 2 = 20

Let A, B and C are the event of selecting of three cans.  The events A, B and C are dependent.

Probability of selecting three cans of juice

P = \frac{\text{number of grape cans}}{\text {total number of cans}}

P (A) = \frac{8}{20}

P (B) = \frac{7}{19}

P (B) = \frac{6}{18}

P = \frac{8}{20} × \frac{7}{19} × \frac{6}{18}

  = \frac{336}{6840}

  = 0.049 or 4.9%

Probability of selecting three cans of grape juice is 4.9%

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To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

P(A|B)=\frac{P(B|A)P(A)}{P(B)}

The probability that a message is spam provided that it contains free is:

P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\

The probability that a message is spam provided that it contains free is approximately 0.789.

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P(A|B)=\frac{P(B|A)P(A)}{P(B)}

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