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Probabilities, really confusing !... Please help

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by praveen_gmat » Wed Oct 12, 2011 6:13 am
Problem 1:
If two dice are thrown simultaneously, what is the probability that one die shows up 2 and the other shows up 5?

ANS:
Total possible outcomes = 36.
Out of these, there are 2 favorable occurrences i.e.,(2,5) and (5,2)
Hence P = 2/36 = 1/18

Problem 2:

When 3 dice are thrown simultaneously, what is the probability that the first die shows up a prime number, second die shows up an odd number and third die shows up an odd prime number?

ANS:
Here
Event A = 2,3,5
Event B = 1,3,5
Event C = 3,5

P(A)=3/6=1/2 ; P(B)=1/2 ; P(C)=1/3

Hence the required probability = 1/2 * 1/2 * 1/3 = 1/12


My question:
In the second problem, the individual probability was found and multiplied.
Why could I have not done the same for problem 1? In which case, the probability would have been 1/6*1/6 = 1/36.

What am I missing? :( Please help.[/u]
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by CappyAA » Wed Oct 12, 2011 6:19 am
I'm not sure if I'm going to use my terms correctly, but the difference is that you are less specific in the first one.

You are asked what is the probability that EITHER die shows up 2 and THE OTHER die shows up 5. So you are multiplying to find the probabilities, but since there are 2 die, you have to add the two probabilities together.

If the question said, what is the probability that the first die shows up 2 and the second die shows up 5, then your answer would be:

1/6*1/6 = 1/36

But because it can happen 2 ways, you add the two probabilities together.

Similarly, if the second one asked what is the probability that one die shows up a prime number, another die shows up an odd number, and a third die shows up an odd prime number (notice how it's not as specific as to which die it could be), you would have to find those individual probabilities and add all the instances together.

Hope that helps!
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by praveen_gmat » Wed Oct 12, 2011 6:23 am
IN the first problem, even after adding, the result is not what is expected.

And the "and" that you are referring to is in the problem statement. Please check again.
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by sam2304 » Wed Oct 12, 2011 6:28 am
praveen_gmat wrote: In the second problem, the individual probability was found and multiplied.
Why could I have not done the same for problem 1? In which case, the probability would have been 1/6*1/6 = 1/36.

What am I missing? :( Please help.[/u]
The probability is not 1/6 * 1/6 but it is 2/6*1/6 = 1/18.

You have two choices for the first dice 2 and 5 (2/6)
But second is dependent on the first's outcome. (1/6)

Hope i am right but not sure though.
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by CappyAA » Wed Oct 12, 2011 6:32 am
praveen_gmat wrote:IN the first problem, even after adding, the result is not what is expected.

And the "and" that you are referring to is in the problem statement. Please check again.
I don't understand what you mean, the first problem is exactly what is expected after adding. In one instance you roll a 2,5 and in the other instance you roll a 5,2. Each has a 1/36 probability of occuring so it's 1/18. If you simply multiply the two probabilities 1/6 together to get 1/36 you are not counting both instances.

The second problem is different because one die is not dependent on the other die's value. So I can multiply the values together (1/2*1/2*1/3) and get the true probability.
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by CappyAA » Wed Oct 12, 2011 6:33 am
sam2304 wrote:
praveen_gmat wrote: In the second problem, the individual probability was found and multiplied.
Why could I have not done the same for problem 1? In which case, the probability would have been 1/6*1/6 = 1/36.

What am I missing? :( Please help.[/u]
The probability is not 1/6 * 1/6 but it is 2/6*1/6 = 1/18.

You have two choices for the first dice 2 and 5 (2/6)
But second is dependent on the first's outcome. (1/6)

Hope i am right but not sure though.
Thanks sam for saying it better than I could :)
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by praveen_gmat » Wed Oct 12, 2011 6:37 am
Fantastic sam.Thank you so much !
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by Brent@GMATPrepNow » Wed Oct 12, 2011 6:45 am
praveen_gmat wrote: My question:
In the second problem, the individual probability was found and multiplied.
Why could I have not done the same for problem 1? In which case, the probability would have been 1/6*1/6 = 1/36.

What am I missing? :( Please help.[/u]
The first probability can be written as an OR probability.

P(one die is 2 and the other is 5) = P(1st is 2 and 2nd is 5 OR 1st is 5 and 2nd is 2)
= P(1st is 2 and 2nd is 5) + P(1st is 5 and 2nd is 2)

At this point, we can multiply the individual probabilities [e.g., P(1st is 2) = 1/6]

So, we get:
P(1st is 2 and 2nd is 5) + P(1st is 5 and 2nd is 2) = (1/6)(1/6)+(1/6)(1/6)
= 1/36 + 1/36
= 2/36
= 1/18

So, in essence, we are still multiplying the individual probabilities, except in the first question, we need to consider two possible cases (which results in an OR probability).

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by cbaum » Wed Oct 12, 2011 6:54 am
In the first problem, you need to account for the fact that 2 and 5 could show up in two different ways, doesn't matter which die each shows up on we just need both numbers. We ARE multiplying to get 1/18, we just need to add to get there (1/6*1/6 + 1/6*1/6) since these are two distinct probabilities.

In the second problem, we're looking for different things on each SPECIFIC die, so we only multiply the probabilities. If we were looking for the number of ways that these different things could happen between the three dice, not specifying which die was supposed to show what, then we would add 1/12+1/12+1/12 (at least this is how I understand it, someone please correct me if I'm wrong)

Hope that makes sense!
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