TIFR CSE 2013 | Part A | Question: 14

Question

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

• A. $\dfrac{1}{(2n)}$
• B. $\dfrac{1}{[(6n)!]}$
• C. $1 - 6^{-n}$
• D. $6^{-n}$
• E. None of the above

Answer 1

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: $E$

Comments

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

Answer 2

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

Answer 3

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 - 1/2ⁿ 

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

Comments

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

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@ Dileep kumar M 6

Probability of getting Odd number is (3/6)^n, not (1/2)^n. Our sample space is  {1,2,3,4,5,6} in this there are 3 odd numbers. So probability of getting odd number in single throw is 3/6

Answer 4

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