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

1. $1/(2n)$
2. $1/[(6n)!]$
3. $1 - 6^{-n}$
4. $6^{-n}$
5. None of the above.

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- $nCn*(\frac{3}{6})^0*(\frac{3}{6})^n$
= 1- $(\frac{1}{2})^n$
So i think option E

answered by Veteran (15.3k points) 17 51 129
selected

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

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.

answered by Boss (8.8k points) 3 8 12
edited by