Question

A basket contains five apples, five peaches, and four pears. You randomly select and eat three pieces of fruit. The first piece of fruit is an apple and the next two pieces are peaches. Find the probability of this occurring.

Answer 1

Answer:

Hence, the answer is:

             $Probability=\dfrac{25}{546}$

Step-by-step explanation:

It is given that:

A basket contains five apples, five peaches, and four pears.

This means that the total number of fruits in the basket are: 5+5+4=14

You randomly select and eat three pieces of fruit.

Now, we are asked to find the probability of the occurrence that an apple is selected first and then two peaches are selected.

i.e. Probability of occurrence=Probability of  drawing an apple×Probability of  drawing two peaches

Hence,

$Probability=\dfrac{5_C_1}{14_C_1}\times \dfrac{5_C_2}{13_C_2}\\\\\\Probability=\dfrac{5}{14}\times \dfrac{5\times 4}{13\times 12}\\\\\\Probability=\dfrac{5\times 5\times 4}{14\times 13\times 12}\\\\\\Probability=\dfrac{25}{546}$

                                             

Answer 2

Total number of fruit = 5 + 5 + 4 = 14
P(A) = Probability of picking an apple on first pick = 5/14
P(B) = Probability of picking a peach on second pick = 5/13
P(C) = Probability of picking a peach on third pick = 4/12
All we have to do is multiply these probabilities:
5/14 * 5/13 * 4/12 = 25/546