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At a food bank a case of donated items contains 10 cans of soup, 8 cans of vegetables, and 8 cans of fruit. If 3 cans are selected at random to distribute, find the probability of getting 1 vegetable and 2 cans of fruit.

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Answer:

0.1385 or 13.85%

Step-by-step explanation:

Number of cans of soup (S) = 10

Number of cans of vegetables (V) = 8

Number of cans of fruit (F) = 8

Total number of cans (n) = 26

There are three possible ways to get 1 vegetable and 2 cans of fruit (VFF, FVF, FFV). If each outcome is equally likely, the probability is:

[tex]P = 3*P(VFF)\\P=3*\frac{8}{26} *\frac{10}{25} *\frac{9}{24}\\P=0.1385[/tex]

The probability is 0.1385 or 13.85%.

Using the combination formula, it is found that there is a 0.0862 = 8.62% probability of getting 1 vegetable and 2 cans of fruit.

  • A probability is the number of desired outcomes divided by the number of total outcomes.
  • In this problem, the order in which the items are chosen is not important, hence the combination formula is used to solve this question.

Combination formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Desired outcomes:

  • 1 vegetable from a set of 8.
  • 2 cans of fruit from a set of 8.

Hence:

[tex]D = C_{8,1} \times C_{8,2} = \frac{8!}{1!7!} \times \frac{8!}{2!6!} = 224[/tex]

Total outcomes:

  • 3 cans from a set of 10 + 8 + 8 = 26.

Hence:

[tex]T = C_{26,3} = \frac{26!}{3!23!} = 2600[/tex]

Then, the probability is:

[tex]p = \frac{D}{T} = \frac{224}{2600} = 0.0862[/tex]

0.0862 = 8.62% probability of getting 1 vegetable and 2 cans of fruit.

To learn more about the combination formula and probability, you can take a look at https://brainly.com/question/24650047

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