286 views

An unbiased die is thrown $n$ times. The probability that the product of numbers would be even is

1. $\dfrac{1}{(2n)}$
2. $\dfrac{1}{[(6n)!]}$
3. $1 - 6^{-n}$
4. $6^{-n}$
5. None of the above.
edited | 286 views

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

$= 1- {^n}C{_n}\times \left(\dfrac{3}{6}\right)^{0}\times \left(\dfrac{3}{6}\right)^{n}$
$= 1- \left(\dfrac{1}{2}\right)^{n}$
So i think option E
edited
0

Prob. of product being even = Favorable outcome/Total outcomes = (6n - 3n)/6n= 1- 1/2n

Total = 6n

Favor. = 6n - [no. formed by only 1,3,5] =6n - 3n

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

reshown

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
0
only one possibility product will be odd when all the numbers is odd.

total possiblities - 6^n

prob that will be odd= 1/6^n

so prob that even will be 1- prob that will be odd = 1-6^-n so answer c.. what is wrong with my approch