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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.

My attempt :

we have $3$ even number and $3$ odd number in the an unbiased dice , i.e. $2,4,6$ and $1,3,5$ respectively .

Probability to occur odd number , i.e. P(odd) $= \cfrac{1}{2}$

As we know , if all numbers are odd then product of these number will be odd , else even , in other words , if at least one even number occur then product of these number will be even .

Therefore , required probability is ,

$=$ probability for at least one even number occur

$= 1 -$ probability for all odd number occur

$= 1 - \left(\cfrac{1}{2}\right)^n$


Product of numbers is even , when an unbiased die rolled ?

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  • $\begingroup$ Yes, you are correct. $\endgroup$ – cr001 Nov 5 '15 at 7:09
  • $\begingroup$ @cr001 dear , I need an verification (regarding the post). $\endgroup$ – 1 0 Nov 5 '15 at 7:39
  • $\begingroup$ Your solution $1-{1\over2}^n$ is correct, what more do you want? $\endgroup$ – cr001 Nov 5 '15 at 7:44
  • $\begingroup$ The answer is none of the above. $\endgroup$ – marshal craft Nov 5 '15 at 7:50
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Your answer is correct.

There's not much left to do with it, except perhaps writing it down slightly more "tidy".


Calculate $1$ minus the probability of the complementary event:

  • The probability of getting an odd number in a roll is $\frac36$
  • The probability of getting an odd product in $n$ rolls is $\left(\frac36\right)^n$
  • The probability of getting an even product in $n$ rolls is $1-\left(\frac36\right)^n$
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  • 1
    $\begingroup$ @MithleshUpadhyay: You're welcome. $\endgroup$ – barak manos Nov 5 '15 at 7:59

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