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A speaks the truth 3 times out of 4, B speaks the truth 7 times out of

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A speaks the truth 3 times out of 4, B speaks the truth 7 times out of  [#permalink]

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New post 03 Sep 2018, 05:13
2
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

68% (01:40) correct 32% (02:21) wrong based on 39 sessions

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A speaks the truth 3 times out of 4, B speaks the truth 7 times out of 10. They both assert that a white ball is drawn from a bag containing 6 balls, all of different colors. What is the probability of the truth of the assertion.
\(a. \frac{12}{49}\)
\(b. \frac{3}{10}\)
\(c. \frac{21}{40}\)
\(d. \frac{3}{40}\)
e. None of these
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Re: A speaks the truth 3 times out of 4, B speaks the truth 7 times out of  [#permalink]

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New post 03 Sep 2018, 10:26
I got 37/40 as the answer. my method of solving:
1. A speaks the truth and B does not speak the truth
2. B speaks the truth and A does not speak the truth
3. Both A and B speak the truth.
Finally add all the 3 priorities.
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Re: A speaks the truth 3 times out of 4, B speaks the truth 7 times out of  [#permalink]

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New post 03 Sep 2018, 10:57
Since it has asked about the assertion of both A and B

A∩B = P(A)* P(B)

Hence the correct answer should be =3/4*7/10 = 21/40

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Re: A speaks the truth 3 times out of 4, B speaks the truth 7 times out of  [#permalink]

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New post 03 Sep 2018, 11:37
1
MithilaGauri wrote:
I got 37/40 as the answer. my method of solving:
1. A speaks the truth and B does not speak the truth
2. B speaks the truth and A does not speak the truth
3. Both A and B speak the truth.
Finally add all the 3 priorities.

Your 1st and 2nd case is impossible.
Let's say the ball is white. If A is telling telling the truth , ball = white.
Now if B is telling a lie , ball is not white. So they keep contradicting each other in an infinite loop.

There's only 2 possible outcomes of this scenario. Either both of them are telling the truth or both of them are lying
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Re: A speaks the truth 3 times out of 4, B speaks the truth 7 times out of   [#permalink] 03 Sep 2018, 11:37
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