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Best answer
21 votes
21 votes
Even number =$ 2,4,6$
odd number = $1,3,5$
Product will come even when even one time even number comes
odd product will be if every time odd comes

so, $P(Even) = 1- P(odd)$

$= 1- {^n}C{_n}\times \left(\dfrac{3}{6}\right)^{0}\times \left(\dfrac{3}{6}\right)^{n}$
$= 1- \left(\dfrac{1}{2}\right)^{n}$

Correct Answer: $E$
edited by
9 votes
9 votes

Here we need to find the probability that at least one time even number should come.

Product of Even and Odd = Even

Product of Even and Even =Even

Product of Odd and Odd =Odd

=1 - probability of coming only odds in "n" throws 

=1-(1/2)n

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4 votes
4 votes

The product will come even when at least one time an even number comes in n rolls.

P(even in 1 roll) =P(odd in 1 roll) =3/6 =1/2 

P(atleast one time even in n rolls) = 1 - P(all times odd in n rolls) = 1 - 1/2n

Hence ,the correct answer is 1 - 1/2

Option (e) None of the above ,is the ans.

edited by
2 votes
2 votes

Odd Numbers = $1,3,5$

Even Numbers = $2,4,6$

Product will be odd iff all n throws result in an odd number

$P(\text{Even product}) = 1 - P(\text{Odd product}) $

There are 3 Odd numbers, so for $n$ throws there are $3^n$ ways in which product will result in an odd number.

$P(\text{Odd product}) =$$\Large \frac{3^n}{6^n} $

$P(\text{Even product}) = 1 - \Large \frac{3^n}{6^n}   = \Large \frac{6^n - 3^n}{6^n}  = \Large \frac{2^n . 3^n - 3^n}{2^n.3^n} =  \Large \frac{3^n( 2^n - 1)}{2^n.3^n} = \Large \frac{( 2^n - 1)}{2^n} =\large 1 - \Large \frac{1}{2^n}   $

e.  None of the above

Answer:

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