Question

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.

Answer 1

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:

$P = 3*P(VFF)\\P=3*\frac{8}{26} *\frac{10}{25} *\frac{9}{24}\\P=0.1385$

The probability is 0.1385 or 13.85%.