# Ten witnesses, each of whom makes but one false statement in six, agree in asserting that a certain event took place, show that the odds are five to one in favour of the truth of their statement, even although the a priori probability of the event is as small as $\dfrac{1}{{{5^9} + 1}}$

Answer

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Hint: Use the probability and events concepts to solve the problem in probability.

Let p be the probability of the event

Then the probability that their statement is true is to the probability that is as false and we know that witnesses=$10$

$ \Rightarrow \dfrac{{{{\left( {\frac{5}{6}} \right)}^{10}}p}}{{(1 - p){{\left( {\frac{1}{6}} \right)}^{10}}}}$

Here we have five odd statements in which to one favour of truth

Total statements =$6$

Odd statements=$5$

So here $\dfrac{{{5^{10}}p}}{{1 - p}}$ represents odd statements in favour of the event

Now in order that the odds in favour of the event may be at least five to one

Then condition will be:

$\frac{{{5^{10}}p}}{{1 - p}} \geqslant 5$

$\

\Rightarrow {5^{10}}p \geqslant 5 - 5p \\

\Rightarrow {5^9}p \geqslant 1 - p \\

\Rightarrow {5^9} + p \geqslant 1 \\

\Rightarrow p({5^9} + 1) \geqslant 1 \\

\Rightarrow p \geqslant \dfrac{1}{{{5^9} + 1}} \\

\ $

Hence we can say that for five odd statements to which one is favour of truth then the prior probability of the event is small as $\dfrac{1}{{{5^9} + 1}}$

NOTE: In this problem total witness will be ignored only by concentrating on the event with 5 odd statements to which one is favour of truth.

Let p be the probability of the event

Then the probability that their statement is true is to the probability that is as false and we know that witnesses=$10$

$ \Rightarrow \dfrac{{{{\left( {\frac{5}{6}} \right)}^{10}}p}}{{(1 - p){{\left( {\frac{1}{6}} \right)}^{10}}}}$

Here we have five odd statements in which to one favour of truth

Total statements =$6$

Odd statements=$5$

So here $\dfrac{{{5^{10}}p}}{{1 - p}}$ represents odd statements in favour of the event

Now in order that the odds in favour of the event may be at least five to one

Then condition will be:

$\frac{{{5^{10}}p}}{{1 - p}} \geqslant 5$

$\

\Rightarrow {5^{10}}p \geqslant 5 - 5p \\

\Rightarrow {5^9}p \geqslant 1 - p \\

\Rightarrow {5^9} + p \geqslant 1 \\

\Rightarrow p({5^9} + 1) \geqslant 1 \\

\Rightarrow p \geqslant \dfrac{1}{{{5^9} + 1}} \\

\ $

Hence we can say that for five odd statements to which one is favour of truth then the prior probability of the event is small as $\dfrac{1}{{{5^9} + 1}}$

NOTE: In this problem total witness will be ignored only by concentrating on the event with 5 odd statements to which one is favour of truth.

Last updated date: 16th Sep 2023

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