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25 Jun 2022, 09:45
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
45% (01:44) correct
55%
(01:42)
wrong
based on 256
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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. \((1-\frac{35}{36})^n\)
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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18 Sep 2024, 07:31
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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
General Discussion
25 Jun 2022, 10:04
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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.
