Question

Two dice are thrown simultaneously. Given that sum of the numbers is NOT more than 5, what is the probability that sum is more than 3?

Answer 1

(A) 7/10, (B) 3/10, (C) 5/6, (D) 3/4, (E) 9/10

Answer 2

This problem involves "conditional probability."
Especially if you're doing this kind of problem for the first time, you might want to write out all 36 possible outcomes of the toss of two dice.

For example: 1-1, 1-2, 1-3,....
2-1, 2-2, 2-3, 2-4,
.....
6-6

Again, if you write these out, or if you draw a "tree diagram," you should obtain 36 possible outcomes.

Next, count how many of these outcomes add upto 5 or less. (That mean eliminating 2-4, 3-5, and so on, both of which add up to more than 5.)

Lastly, count how many of the outcomes from the set you just created, count how many add up to more than 3 (for example: 1-3, 2-3, and so on.

The answer you're looking for is this:

P(More than 3 | not more than 5)=
Count of how many are More than 3
--------------------------------------------
count of how many (total) are not more than 5

If this is not clear enough to enable you to find your own answer, show what you have done and I will gladly return and help you solve this problem.

Answer 3

Let \(X\) be the sum of the rolls

you need

\[P(X>3|X\le5)=\frac{P(X>3\text{ and }X\le5)}{P(X\le 5)}=\frac{P(3<X\le5)}{P(X\le 5)}=\cdots\]

Answer 4

ok i got 7/10. There where 10 combinations that added up to five or less. within those ten there were 7 that added up to more than 3. Was i close?

Answer 5

@mathmale

Answer 6

I would need to do this whole problem myself before responding to your "was i close?"

As expected, there are a total of 36 possible outcomes. I've written them all out.

Eliminating all combinations that add up to more than 5, I find myself left with 1-1, 1-2, 1-3, 1-4, 2-1, 2-2, 2-3, 3-1, 3-2, 4-1. How many there? 10.

How many of these 10 add up to more than 3? I'm a bit lazy, so I count how many add up to 3 or less, and thus obtain 1-1, 1-2, 2-1, or 3.

Thus, you are correct in concluding that out of the ten that do NOT add up to more than 5, seven DO add up to 3 or less.

So, your 7/10, or 0.7, is

Answer 7

RIGHT ON TARGET! NICE WORK!!

Answer 8

Please note that the equation written by Zarkon, earlier,\[P(X>3|X\le5)=\frac{P(X>3\text{ and }X\le5)}{P(X\le 5)}=\frac{P(3<X\le5)}{P(X\le 5)}=\cdots\]

Answer 9

is the formal way to do this problem and is well worth studying and knowing. His formula is for CONDITIONAL PROBABILITY.

Answer 10

Thank you @mathmale and @zarkon i appreciate it.

Answer 11

You're very welcome. You appear to be doing very well.