Question

A basket contains four apples, three 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

Is the answer 1/48
possible outcomes: 4*3*4 = 12*4 = 48
Each outcome is different that means that this can only happen once
one out of forty eight or 1/48

Answer 2

Answer: $\dfrac{1}{30}$

Step-by-step explanation:

Given : Number of apples in basket = 4

Number of peaches in basket = 3

Number of pears in basket = 4

If we randomly select and eat three pieces of fruit.

The number of ways to select an apple and the next two pieces are peaches :_

$^4P_1\times ^3P_2\\\\=\dfrac{4!}{(4-1)!}\times\dfrac{3!}{(3-2)!}=4\times6=24$

The total number of ways to select 3 fruits from 10 fruits :-

$^{10}P{3}=\dfrac{10!}{(10-3)!}=10\times9\times8=720$

Now, the probability that the first piece of fruit is an apple and the next two pieces are peaches will be :-

$\dfrac{24}{720}=\dfrac{1}{30}$