How many outcomes does this experiment have?

Question

For this problem, assume the balls in the box are numbered 1 through 7, and that an experiment consists of randomly selecting 2 balls one after another without replacement. 

(1) How many outcomes does this experiment have?

(2) What probability should be assigned to each outcome?

(3) What probability should be assigned to the event that at least one ball has an odd number?

Answer 1

Hey Katie,

 

1) You have 7 choices drawing the first ball and 6 choices drawing the second, so the total number of outcomes is 42 when selecting two balls without replacement.

 

2) Each outcome is equally likely -- assume you're not peeking! Thus each outcome has a probability of 1 in 42, which is 1/42 = 0.0238.

 

3) First, let's decide if order counts in this problem. If you draw a 3 and then a 4, is that distinct from drawing a 4 then a 3? After all, you're holding a 3 and a 4 in your hand when you're done -- does it matter how you got them?

 

Yes, because there are two ways of ending up with a 3 and a 4 in your hand!

 

Before calculating this out, estimate your answer, so you'll know whether or not your calculation is way off later. Since more than half the balls are odd, the odds of getting one odd ball should in the first draw will be better than 50%, right?  So the chances of getting at least one odd ball in two draws should approach 100%. Just as a rough estimate, this will work. On we go.

 

Since order counts, we need to add three kinds of outcomes to get this one right: a) the first ball is odd but the second one isn't, b) the second ball is odd but the first one isn't, and c) both balls are odd.

 

BUT WAIT!Look at the complement of the answer you want: the only other possible outcome for drawing two balls is

 

3d) when both balls are even, right? So let's just count those outcomes, and then use 42 minus that number to calculate the number of ways to get at least one odd ball.

 

The only way of getting two balls even are 2 and 4, 2 and 6, 4 and 2, 4 and 6, 6 and 2, 6 and 4. That's 6 ways. Thus there are 36 ways to draw at least one odd ball in your two draws.

 

Thus the probability of drawing at least one odd ball in two draws is 36/42 = 6/7 = about 0.86.

 

This passes what I call the "sniff test," which compares this calculated answer to our best guess of some number approaching 100%.

 

We're done, and with a high degree of confidence!

 

 

Cheers!

 

Eric M.

McMinnville OR