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On a lunch counter, there are 3 oranges, 5 apples, and 2 bananas. If 3 pieces of fruit are selected, find the probability that 1 orange, 1 apple, and 1 banana are selected (does not have to be in that order).

Answer :

MrRoyal

Answer:

[tex]Probability = \frac{1}{24}[/tex]

Step-by-step explanation:

Given

[tex]Oranges = 3[/tex]

[tex]Apples = 5[/tex]

[tex]Banana = 2[/tex]

[tex]Total = 10[/tex]

Required

Determine the probability of 1 orange, 1 apple and 1 banana

Since, order is not important:

[tex]Probability = P(Orange) * P(Apple) * P(Banana)[/tex]

[tex]Probability = \frac{3}{10} * \frac{5}{9} * \frac{2}{8}[/tex]

The difference in the numerator is as a result of picking the fruit without replacement

[tex]Probability = \frac{30}{720}[/tex]

[tex]Probability = \frac{1}{24}[/tex]

Probability is simply how likely something is to happen.

The probability of 1 orange, 1 apple, and 1 banana are  [tex]\frac{1}{24}[/tex]

On lunch counter,

Number of oranges = 3

Number of apples = 5

Number of bananas = 2

If 3 piece of fruit is selected then, probability of 1 orange, 1 apple and 1 banana

                          [tex]=\frac{3}{10}*\frac{5}{9} *\frac{2}{8} \\\\=\frac{30}{720}\\\\=\frac{1}{24}[/tex]

Therefore, the probability of  selecting 1 orange, 1 apple, and 1 banana are [tex]\frac{1}{24}[/tex]

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