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There are two dice, each numbered 1 to 6. If the dice are thrown n times, what is the probability of obtaining at least one double "4" in the n throws ?
A. \(1-(\frac{1}{36})^n\)
B. \(1-(\frac{25}{36})*n\)
C. \(1-(\frac{25}{36})^n\)
D. \(1-(\frac{35}{36})^n\)
E. \((1-\frac{35}{36})^n\)
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A. \(1-(\frac{1}{36})^n\)
B. \(1-(\frac{25}{36})*n\)
C. \(1-(\frac{25}{36})^n\)
D. \(1-(\frac{35}{36})^n\)
E. \((1-\frac{35}{36})^n\)
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Official Solution:
There are two fair dice, each numbered 1 to 6. If the dice are thrown \(n\) times, what is the probability of obtaining at least one double "4" in the \(n\) throws?
A. \(1-(\frac{1}{36})^n\)
B. \(1-(\frac{25}{36})*n\)
C. \(1-(\frac{25}{36})^n\)
D. \(1-(\frac{35}{36})^n\)
E. \((1-\frac{35}{36})^n\)
The probability of getting a double "4" is \(\frac{1}{6}*\frac{1}{6}=\frac{1}{36}\).
The probability of NOT getting a double "4" is \(1 - \frac{1}{36}=\frac{35}{36}\).
The probability of NOT getting a double "4" in \(n\) throws is \((\frac{35}{36})^n\).
Therefore, the probability of obtaining at least one double "4" in the \(n\) throws is \(1 - (\frac{35}{36})^n\).
Answer: D
_________________
There are two fair dice, each numbered 1 to 6. If the dice are thrown \(n\) times, what is the probability of obtaining at least one double "4" in the \(n\) throws?
A. \(1-(\frac{1}{36})^n\)
B. \(1-(\frac{25}{36})*n\)
C. \(1-(\frac{25}{36})^n\)
D. \(1-(\frac{35}{36})^n\)
E. \((1-\frac{35}{36})^n\)
The probability of getting a double "4" is \(\frac{1}{6}*\frac{1}{6}=\frac{1}{36}\).
The probability of NOT getting a double "4" is \(1 - \frac{1}{36}=\frac{35}{36}\).
The probability of NOT getting a double "4" in \(n\) throws is \((\frac{35}{36})^n\).
Therefore, the probability of obtaining at least one double "4" in the \(n\) throws is \(1 - (\frac{35}{36})^n\).
Answer: D
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General Discussion
Posts: 1,070
Bunuel
The probability of double4 on any given throw is 1/6 * 1/6 = 1/36. The probability of notadouble4 = 35/36.
If n=2, we need 1-(notadouble4 AND notadoubel4). \(1-(\frac{35}{36})^2\)
Only D fits.
Answer choice D.
Very intelligent question where we attack what the author doesn’t want and subtract from the sample set.
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Why the probability of having a double 4 isn't (5/6)*(5/6)?
I understand it logically but not mathematically.
I understand it logically but not mathematically.
Nasco
Probability of having first dice 4 is 1/6
Probability of having second dice 4 is 1/6
Then we have to use AND probability...
Probability of having two dice 4 is: 1/6*1/6
Therefore answer is 1/36.
Now probability of having not having double 4 is: 1-1/36= 35/36
Therefore answer.. Atleast one double 4= 1-(probably of having no double 4)
Thus: 1-35/36
For n times... 1- (35/36)^n
D
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Nasco
I think you're asking why the probability of NOT having a double 4 isn't (5/6)*(5/6).
(5/6)(5/6) represents the probability that neither die shows a 4, meaning both dice must show numbers other than 4. However, not having a double 4 means any combination where at least one die is not a 4, so any combination except (4, 4) is allowed. Combinations like (1, 4) or (4, 3) are possible, which wouldn't be included in (5/6)(5/6).
Hope that makes it clearer!
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Bunuel
Thank you for explaining
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