Question

a dice is thrown 2n+1 times. the probability that faces with even numbers show odd number of times is

Answer 1

Look on Google and search on it

Answer 2

Answer:

1/2

Step-by-step explanation:

The number of throws is odd (2n+1 is an odd number).

Let a be the number of even numbers and b the number of odd numbers.

Then a+b = 2n+1, so a being odd is the same thing as b being even.

That is:

   P(a is odd) = P(b is even).            ... (1)

We want to find P(a is odd).

Also, in any one throw:

  P(even number) = 3/6 = P(odd number).

So getting an even number of even numbers has the same probability as getting an even number of odd numbers.  That is:

    P(a is even) = P(b is even)            ... (2)

Now...

P(a is odd) + P(a is even) = 1           [ since one of these must occur ]

=>  P(a is odd) + P(b is even) = 1      [ by (2) ]

=>  P(a is odd) + P(a is odd) = 1        [ by (1) ]

=> 2 P(a is odd) = 1

=> P(a is odd) = 1/2