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Bunuel
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There are two dice, each numbered 1 to 6. If the dice are thrown n tim 26 Feb 2021, 07:30
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34% (02:00) correct 66% (02:41) wrong based on 62 sessions

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There are two dice, each numbered 1 to 6. If the dice are thrown n times, what is the determine the least value of n for which the probability of at least one double 4 is greater than 1/2?

A. 24
B. 25
C. 26
D. 27
E. 28


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There are two dice, each numbered 1 to 6. If the dice are thrown n tim 26 Feb 2021, 08:17
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The chance of getting double 4s is 1/36. In order to have 1/2 probability we need to raise (35/36) ^ x and calculate when x is big enough for this number to reach the first answer equal to or less than one half.
The issue with this question is that this will not be a possible gmat question, given that there is no calculator, and the answer choices given by the question are very similar given those results i feel somebody should be an expert in logs to calculate this. (Personal opinion)
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There are two dice, each numbered 1 to 6. If the dice are thrown n tim 12 Mar 2021, 23:41
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There are two dice, each numbered 1 to 6. If the dice are thrown n times, what is the determine the least value of n for which the probability of at least one double 4 is greater than 1/2?

P(getting a double 4) = 1/6*1/6 = 1/36; P (not getting a double 4) = 1-1/36 = 35/36

Given, at least a double should be greater than 1/2.
So, Not getting a double 4 in n throws will be less than 1/2.

or, (35/36)^n < 1/2 or, n~25 (at n= 25, Prob. will be 0.494)

So, I think B. :)
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There are two dice, each numbered 1 to 6. If the dice are thrown n tim 18 Mar 2025, 00:46
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Step 1: Probability of Not Rolling (4,4)
The probability of rolling (4,4) in a single roll is:
\(P(4,4) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)
The probability of not rolling (4,4) in a single roll is:
\(1 - \frac{1}{36} = \frac{35}{36}\)
If you roll the dice \(n\) times, the probability of never rolling (4,4) is:
\(\left(\frac{35}{36}\right)^n\)
Step 2: Finding the Smallest \(n\)
We need to find the smallest \(n\) such that the probability of at least one (4,4) is greater than \(\frac{1}{2}\). This means:
\(1 - \left(\frac{35}{36}\right)^n > \frac{1}{2}\)
Rearranging:
\(\left(\frac{35}{36}\right)^n < \frac{1}{2}\)
Step 3: Estimating \(n\) Without a Calculator
We can approximate it using the exponential approximation:
\((1 - x)^n \approx e^{-nx} \quad \text{for small } x.\)
Here, \(x = \frac{1}{36}\), so:
\( \left(\frac{35}{36}\right)^n \approx e^{-n/36}\)
We need:
\(e^{-n/36} < \frac{1}{2}\)
Since a well-known approximation is:
\(e^{-0.7} \approx 0.5,\)
we can set:
\(\frac{n}{36} \approx 0.7\)
Solving for \(n\):
\(n \approx 0.7 \times 36 = 25.2\)
Since \(n\) must be an integer, we round up to \(n = 25\).

Bunuel
There are two dice, each numbered 1 to 6. If the dice are thrown n times, what is the determine the least value of n for which the probability of at least one double 4 is greater than 1/2?

A. 24
B. 25
C. 26
D. 27
E. 28


Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
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