Aptitude Discussion

Q. |
$A$ speaks truth in 75% of cases and $B$ in 80% of cases. In what percent of cases are they likely to contradict each other in narrating the same event? |

✔ A. |
35% |

✖ B. |
5% |

✖ C. |
45% |

✖ D. |
22.5% |

**Solution:**

Option(**A**) is correct

Different possible cases of contradiction,

$A$ speaks truth and $B$ does not speaks truth.

Or, $A$ does not speak truth and $B$ speaks truth.

$= (3/4 × 1/5) + (1/4 × 4/5)$

$= 3/20 + 4/20$

$= 7/20$

$= \textbf{35%}$

**Anonymous**

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**Ashher**

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I think this is the case of "DOUBLE EVENT PROBABILITY".

The probability that two mutually exclusive events will have the same specific occurrence is the product of their probabilities.

Reply of Riya...

**Sunny**

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what if both are telling lies

$P(A'nB')=P(A')\times P(B')$

$=\dfrac{1}{4} \times \dfrac{1}{5}$

$=\dfrac{1}{20}$

$=.05$

$=5\%$

Ashher, I don't think you have solved it right.

For, $P(A'nB')=P(A')\times P(B')$, Events $A$ and $B$ have to be mutually exclusive. And how do you ensure that?

I think this is the case of "DOUBLE EVENT PROBABILITY".

The probability that two mutually exclusive events will have the same specific occurrence is the product of their probabilities.

But if both are telling lies then how is it MUTUALLY EXCLUSIVE event?

they dont have any common... A is other person and B is another person... Like 2 dies or 2 coin...

I guess you are mistaken about the concept of mutually exclusive events. Different events aren't mutually exclusive they need to be INDEPENDENT OF EACH OTHER i.e. one event does not affect the other.

P(both are telling the true) =75%x80%=60%

P(both are telling lie) =25%x20%=5%

P(there is no contradiction)=5%+60%=65%

P(contradiction)=1-(65%)=35%