Problem Solving

It's a simple one but I'm not sure what the best approach is.

If there are six numbers numbered 1-6, what is the probability of choosing numbers 1, 2, and 3?


My approach: because order doesn't matter I used the combination 6 choose 3 which equals 20 possible combinations. Then there is only one combination out of these 20 that has 1, 2, and 3 so the probability is 1/20. Is that right?

If anyone can post with an explanation that would be fantastic.

the general rule for probablity is:

number of desired outcomes
---------------------------------
number of total possible outcomes

I assume that you are asking for the probability of getting a 1 or a 2 or a 3.

In this case, the number of desired outcomes is 3 (for values 1 or 2 or 3) and the number of total possible outcomes is 6.

So, the solution is 3/6 = 1/2.

Another approach, is to calculate the probability of each event and add them.
probability of getting 1 = probability of getting 2 = probability of getting 3 = 1/6.

1/6 + 1/6 + 1/6 = 3/6 = 1/2

Hope it helps!

uymba wrote:the general rule for probablity is:

number of desired outcomes
---------------------------------
number of total possible outcomes

I assume that you are asking for the probability of getting a 1 or a 2 or a 3.

In this case, the number of desired outcomes is 3 (for values 1 or 2 or 3) and the number of total possible outcomes is 6.

So, the solution is 3/6 = 1/2.

Another approach, is to calculate the probability of each event and add them.
probability of getting 1 = probability of getting 2 = probability of getting 3 = 1/6.

1/6 + 1/6 + 1/6 = 3/6 = 1/2

Hope it helps!

I guess question is 1, 2 AND 3. In this case, is 1/6 * 1/6 * 1/6 the answer?

The Question doesnt say that u are replacing the number after picking,

So P(PICKING 1) = 1/6
P(picking 2) = 1/5
P(picking 3) = 1/4

P(picking 1,2 and 3) is 1/6*5*4 = 1/120

whats the OA?