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Probability bouncer

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Probability bouncer

by louvre » Wed Sep 03, 2008 7:40 am
Q. Two dice, each numbered 1 to 6, are thrown n times. Determine the least value of n for which the probability of at least one double 4 is greater than 1/2.

Hi,
I don't know the authenticity of the above question and I don't have the answers. I don't even know if it qualifies for a GMAT question, but it's important for me to understand the concept of Probability as such.

here is my explanation.

Step 1:
Probability of double 4 (i.e. [4,4]) in 1 attempt is 1/36.
Pr. of atleast 1 occurrence is 1-(Pr. of NO [4,4]) ==> 1-(1/36)=35/36
==> in n attempts it's (35*n)/36.

Step 2:

This Pr. should be greater than 1/2
==> 35n/36 > 1/2
==> 35n > 16

I'm stuck here, IF ( a big IF!!) my approach is correct, how to determine the n?

Thanks in advance for your responses.
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by Ian Stewart » Wed Sep 03, 2008 9:41 am
You started off well:

The probability you do NOT get a double-4 is 35/36

So the probability you do NOT get a double-4 on two rolls is (35/36)*(35/36) = (35/36)^2, and the probability you do not get a double-4 on n rolls is (35/36)^n.

We want the probability of NOT getting double-4 to be less than 1/2, because this will guarantee that the probability of getting a double-4 at least once is greater than 1/2. So we would need to find the smallest value of n that makes the following inequality true:

(35/36)^n < 1/2

That would be a lot of work to calculate, however.
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by hengirl03 » Wed Sep 03, 2008 9:48 am
Can you do the problem like this:

Probability of getting double 4 on first roll: 1/36

18(1/36) = 1/2

So, the least value of n for which the probability of at least one double 4 is greater than 1/2 is 19.
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by Ian Stewart » Wed Sep 03, 2008 9:57 am
Unfortunately, you cannot do the problem in that way. If you try to apply your method to the following problem:

After rolling a pair of dice 36 times, what is the probability that you will get a double-4 at least once?

you would find that the probability is 36(1/36) = 1; that is, you would find that it is absolutely certain that you get a double-4 once if you roll a pair of dice 36 times. Of course, this is not a certainty; theoretically it's possible we roll double-6 every time, for example. By adding the probabilities here, you are assuming that each roll is different from all previous rolls, which we cannot do here.
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by louvre » Thu Sep 04, 2008 11:02 am
Thanks Ian,
I'm starting to understand this topic.
I thought it's n times the probability but how is it nth power?
Thanks again!
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by Ian Stewart » Thu Sep 04, 2008 11:22 am
louvre wrote:Thanks Ian,
I'm starting to understand this topic.
I thought it's n times the probability but how is it nth power?
Thanks again!
It's because we're using what's often called the 'And Rule' in probability. Let's do a simple problem:

What's the probability, when you roll one die seven times, that you get a 5 every time?

The probability of getting a 5 on one roll is 1/6. If we need to get a 5 on the first roll *and* on the second roll, we multiply: (1/6)(1/6). If we need to get a 5 on the first *and* the second *and* the third ... *and* the seventh rolls, we'll multiply (1/6) seven times: we'll get (1/6)^7. If we wanted to know the probability of getting a 5 every time if you roll a die n times, the probability would be (1/6)^n for the same reason.

That's all I was doing in the above question, but the above question was a bit more complicated. We wanted to know the probability we do *not* get a 4-4 on two dice if we roll the two dice n times. On one roll, the probability you do not get 4-4 is 35/36; on two rolls, the probability of not getting 4-4 either time is (35/36)^2, and on n rolls, the probability is (35/36)^n.

Hope that makes things clear!
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by N.O » Tue Oct 21, 2008 12:51 pm
Once you calculate that (35/36)^n<(1/2) , just log both sides

log((35/36)^n)<log(0.5) -->

n*log(35/36)<log(0.5) -->

n<(log0.5)/(log35/36) -->

n<log(0.5/(35/36)) -->

n<log(18/35) --> The answer

Note: GMAT does not require log. It is not a GMAT question!
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