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The probability of Loki getting a question correct on a certain exam

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Abhi077
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The probability of Loki getting a question correct on a certain exam [#permalink] New post   Updated on: 06 Feb 2019, 07:02
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The probability of Loki getting a question correct on a certain exam is 75 percent. If Loki answers three test questions in a row, what is the probability that he gets at least two of the questions wrong?


A) \(\frac{5}{64}\)

B) \(\frac{5}{32}\)

C) \(\frac{3}{16}\)

D) \(\frac{7}{32}\)

E) \(\frac{5}{16}\)
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B

Originally posted by Abhi077 on 06 Feb 2019, 06:59.
Last edited by Bunuel on 06 Feb 2019, 07:02, edited 1 time in total.
Renamed the topic and edited the question.
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Re: The probability of Loki getting a question correct on a certain exam [#permalink] New post   06 Feb 2019, 07:25
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Probability(Right) = 3/4
Probability(Wrong) = 1/4

At least two wrong = RRW+RWR+WRR+RRR = 9/64+1/64 = 10/64 = 5/32

B is the answer.
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Re: The probability of Loki getting a question correct on a certain exam [#permalink] New post   06 Feb 2019, 11:05
Abhi077 wrote:
The probability of Loki getting a question correct on a certain exam is 75 percent. If Loki answers three test questions in a row, what is the probability that he gets at least two of the questions wrong?


A) \(\frac{5}{64}\)

B) \(\frac{5}{32}\)

C) \(\frac{3}{16}\)

D) \(\frac{7}{32}\)

E) \(\frac{5}{16}\)


Correct = 75% = 3/4
Wrong = 1/4

Now at least 2 wrong out of 3

3!/2! (1/4)*(1/4)*(3/4) + 1/64

10/64

5/32

B
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Re: The probability of Loki getting a question correct on a certain exam [#permalink] New post   11 Feb 2019, 07:22
KanishkM wrote:
Abhi077 wrote:
The probability of Loki getting a question correct on a certain exam is 75 percent. If Loki answers three test questions in a row, what is the probability that he gets at least two of the questions wrong?


A) \(\frac{5}{64}\)

B) \(\frac{5}{32}\)

C) \(\frac{3}{16}\)

D) \(\frac{7}{32}\)

E) \(\frac{5}{16}\)


Correct = 75% = 3/4
Wrong = 1/4

Now at least 2 wrong out of 3

3!/2! (1/4)*(1/4)*(3/4) + 1/64

10/64

5/32

B


can you please tell me where you got 1/64 from? I was following till 1/64 showed up haha
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Re: The probability of Loki getting a question correct on a certain exam [#permalink] New post   11 Feb 2019, 07:29
eddie9233 wrote:
KanishkM wrote:
Abhi077 wrote:
The probability of Loki getting a question correct on a certain exam is 75 percent. If Loki answers three test questions in a row, what is the probability that he gets at least two of the questions wrong?

Correct = 75% = 3/4
Wrong = 1/4

Now at least 2 wrong out of 3

3!/2! (1/4)*(1/4)*(3/4) + 1/64

10/64



can you please tell me where you got 1/64 from? I was following till 1/64 showed up haha


Sure eddie9233

So the question is asking for at least 2 wrong,

1/64 is the case when all 3 question are wrong, 1/4 * 1/4 * 1/4
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Re: The probability of Loki getting a question correct on a certain exam [#permalink] New post   12 Feb 2019, 20:56
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Abhi077 wrote:
The probability of Loki getting a question correct on a certain exam is 75 percent. If Loki answers three test questions in a row, what is the probability that he gets at least two of the questions wrong?


A) \(\frac{5}{64}\)

B) \(\frac{5}{32}\)

C) \(\frac{3}{16}\)

D) \(\frac{7}{32}\)

E) \(\frac{5}{16}\)


We need to determine the probability that Loki answers at least two questions wrong, so:

P(C-W-W) or P(W-W-W)

P(C-W-W) = 3/4 x 1/4 x 1/4 = 3/64

Since C-W-W can be arranged in 3!/2! = 3 ways, the probability is 3 x 3/64 = 9/64.

P(W-W-W) = (1/4)^3 = 1/64

Thus, the probability of answering at least two questions wrong is 9/64 + 1/64 = 10/64 = 5/32.

Answer: B
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The probability of Loki getting a question correct on a certain exam [#permalink] New post   13 Sep 2019, 04:43
Abhi077 wrote:
The probability of Loki getting a question correct on a certain exam is 75 percent. If Loki answers three test questions in a row, what is the probability that he gets at least two of the questions wrong?

A) \(\frac{5}{64}\)
B) \(\frac{5}{32}\)
C) \(\frac{3}{16}\)
D) \(\frac{7}{32}\)
E) \(\frac{5}{16}\)


CASES
\(WWR: (1/4)^2(3/4)•(3!/2!)=3/64•3=9/64\) (\((3!/2!)\) is how many ways we can arrange WWR);
\(WWW: (1/4)^3=1/64\);
\(P(WWR+WWW)=10/64=5/32\)

Answer (B)
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gmatclubot
The probability of Loki getting a question correct on a certain exam [#permalink]
13 Sep 2019, 04:43
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