# Probability - I need help understanding the "mechanics"

1. Jul 2, 2014

### christian0710

Hi, Can someone help me understand why what I am doing is wrong? How much of my logic is correct and where do i fail to understand this probability problem? And what am I actually trying to calculate by doing what I do?

If i have 18 bananas and 12 apples and they have the same chance of being picked,
And i have to calculate the probability of getting 2 bananas and 2 apples then my logic tells me,

The probability of picking one banana must be 18/30 (right?) The probability of picking 2 bananas in a row must be 18/30 * 18/30 = 9/25 (right?)
The probability of picking 2 apples in a row must be
12/30*12/30 = 4/25.

The probability of picking 2 bananas and 2 apples can occur in different ways/orders can't be calculated by

(18/30 * 18/30)*(12/30*12/30) . Why is that?

I know that the only true way to calculate this problem, is to first find out how many ways i can pick 4 fruits (number of possible combinations of 4), then use the binomial coefficient to find out in which ways i can pick 2 bananas, then 2 apples and then multiply the numbers together and divide it with the total number of possible combinations. I know this because it's in my textbook, but my intuition would prefere to do it the way i did it above, so i want to understand what I'm doing wrong, and if there are perhaps a explanation to why this is the only way you can calculate this problem?

2. Jul 2, 2014

Remember - as you pick fruits of each type two things occur:
* the number of fruit of that type decreases by 1
* the total number of fruit remaining for further picks has decreased by 1

So, for bananas, you can't have both numerators 18 and both denominators 30

Second: It will be much easier if you consider the problem as requiring you to select "2 bananas from the 18 available" and "2 apples from the 12 available"

3. Jul 2, 2014

### verty

This is picking without replacement: there are 30*29*28*27 ways to pick 4 fruit from 30 without replacement. The task here is to count how many of those picks are successful at picking two bananas and two apples.

There are 18 * 17 * 12 * 11 ways to pick two bananas, followed by two apples. But this is not the only order, you can have BBAA, BABA, BAAB, etc. Can you see how to continue? If there are C of these BBAA letter orders, you have counted 1/C of the successful ways of picking two bananas and two apples.

4. Jul 3, 2014

### Ray Vickson

The probability of picking two bananas first is not (18/30)*(18/30); rather, it is (18/30)*(17/29), because after picking a banana first you are left with 17 bananas and 29 pieces of fruit. After that, the probability of picking two apples next is (12/28)*(11/27) for similar reasons. So, the probability of the specific picks BBAA in that order is (18/30)*(17/29)*(12/28)*(11/27). I will leave it to you to figure out how to get the probability of two bananas and two apples in any order (that is, when no order is specified).