Simple Maths Question.

March 4, 2006 8:56 AM

Simple Maths Question. If two people have an indvidual chance of 1/4 of gaining entry to the NY Marathon, what chance is there that one of them will get in. Both with be successful and neither will be successful. Any explanation would be helpful. Thanks
posted by Cookie Monster to education (8 comments total)

Call the people A and B.

The chance A gets in is 1/4, and the chance B doesn't is 3/4. So the chance both of these events happen is the product of these two: 3/16. You can reverse this situation, so B gets in and A doesn't, and the chance of that is also 3/16. So the chance of any one person (either A or B but not both) getting in is the sum of these two: 6/16, or 3/8.

Similarly, for both to get in, two events of chance 1/4 have to happen, so you multiply them together to get 1/16, and for neither to be successful, two events of chance 3/4 have to happen, so you get 9/16.

To summarise:

One of them gets in: 3/8
Both: 1/16
Neither: 9/16
posted by Orange Goblin at 9:16 AM on March 4, 2006

For this problem, you can imagine it as the result of throwing two four-sided dice.

Each die has only one side that is 'successful' and three sides that are 'unsuccessful'. So, when you throw a die, the probability of success is 1/4, like in your problem.

First, how many total outcomes are there? Imagine you throw one die at a time. There are four ways the first die can land, and, for each of these, four ways the other die can land.

The total number of combinations this gives you is 4*4 = 16.

Now, it's pretty easy to just enumerate all the possible combinations.

There's exactly one where both are 'successful', so the probability of that is:

Both successful: 1/16

For the first die, there are three ways for it to land 'unsuccessful', and for each of these three, there are three ways for the next to land 'unsuccessful', so:

Both unsuccessful: 9/16

What about exactly one die landing 'successful'? There is one way for the first die to land successful, but, given this, there are still three ways for the second die to land 'unsuccessful', so the probability of the first die being 'successful' and the second die being 'unsuccessful' is 3/16. The probability of the opposite event, that the first die is 'unsuccessful', the the second is 'successful', is the same, 3/16. You can add these together, so:

Exactly one successful: 6/16

These are all the cases. Note that they all add up to one:

1/16 + 9/16 + 6/16 = 16/16 = 1
posted by driveler at 9:21 AM on March 4, 2006

To explain slightly more generally: events have probabilities (or chances) of happening. If you want to know the probability of two events happening, as long as they are independent (so in this example, if A had more of a chance of getting in if B had already got in, the events would not be independent), you can multiply the probabilities. Also, if an event can happen in more than one way, you have to add the probabilities of all the different ways to find the full probability of the event occurring, hence the first section of my previous answer.
posted by Orange Goblin at 9:23 AM on March 4, 2006

The probability is 1/16.

One event is dependent on the other, so you can think of it this way: the first person will get in 1 out of every 4 times. For the other 3/4 times, no matter what happens to the second person, it will NOT be the case that both will get in. So in the 1/4 chance that the first person gets in, there is another 1/4 chance that the second person will get in. Hence, 1/4 * 1/4 = 1/16.

Another way to think about it: imagine that you roll two four-sided dice, hoping to get two fours (this is equivalent to your question, and you can think of each roll as "four" or "not four"). There are 16 possible combinations of rolls:

1-1 (no-no)
1-2 (no-no)
1-3 (no-no)
1-4 (no-entry)
2-1 (no-no)
2-2 (no-no)
2-3 (no-no)
2-4 (no-entry)
3-1 (no-no)
3-2 (no-no)
3-3 (no-no)
3-4 (no-entry)
4-1 (entry-no)
4-2 (entry-no)
4-3 (entry-no)
4-4 (entry-entry)

Only ONE of those combinations is two fours, hence the probability is 1/16.

Why do I get the feeling I've just done homework for someone?
posted by The Michael The at 9:24 AM on March 4, 2006

Assuming event A (the first friend getting in) and event B (the second friend getting in) are independent, the answer is (probability of A) x (probability of B) or 1/4 x 1/4 = 1/16

...basic probability. Check any textbook.
posted by bim at 9:37 AM on March 4, 2006

Thanks, people.

6.25% I wont start training just yet then.
posted by Cookie Monster at 9:49 AM on March 4, 2006

You can also demonstrate this visually, if that's helpful to you. Using "1" to indicate success , and "0" to indicate failure, we have

Person A: 1 0 0 0
Person B: 1 0 0 0

Now if we put the two persons on to orthogonal Cartesian axes, with the sum of people who get in at each of the 16 intersection, we get


posted by orthogonality at 11:11 AM on March 4, 2006

The phrase "one of them will get in" is a little ambiguous. If you mean one and only one, that's a condition know as an eXclusive OR (XOR). If you mean at least one gets in, that's a simple OR condition.

Orange Goblin's and driveler's calculations and explanations are correct for the XOR case (as well as the other two cases of both and none). The probability of the OR condition can be computed several ways: add the probability of the XOR to the probability of the Both (3/8 + 1/16 = 7/16), subtract the probability of the None from 1 (1 - 9/16 = 7/16) or use boolean math with DeMorgan's theorem to construct an OR gate from an AND gate (NOT them, AND them, and NOT them again) to get [1-(1-1/4)(1-1/4)] = 7/16.

BTW, I think it's cool to see the different approaches to the answer here. I've been doing the boolean math thing for so long (it's easier to plug and chug) that I'd almost forgotten there were other methods that are much easier to explain. Thanks, guys.
posted by forrest at 1:55 PM on March 4, 2006