Question

A die is tossed twice. Getting an odd number in at least a toss is termed as a success. Find the
probability distribution of number of successes. Also find expected number of successes.

Answer 1

1/4 is answer of this question

Dice = {1,2,3,4,5,6}
number of throw =2 means
odd no in dice={1,3,5}
probablity for first throw = n(e)/n(s)=3/6=1/2
probablity for second throw = n(e)/n(s)=3/6=1/2
answer is 1/2*1/2=1/4

Answer 2

Getting odd number in one throw or toss:    probability = 3/ 6 = 1/2
probability of getting even number:  1/2
Let n be the number of  odd numbers obtained when two dice are thrown  or one die is thrown tiwice.

probability of getting zero odd numbers
    = P(even number AND even number in 2nd throw)
 P( n = 0)    = 1/2 * 1/2 = 1/4

Probability of getting one odd number = P(n = 1)
    = P (Odd number AND even number) + P(even number AND odd number)
   = 1/2 * 1/2 + 1/2 * 1/2
   P(n = 1) = 1/2

Probability of getting two odd numbers =
   P( n = 2) =  P( odd number AND odd number)
   =  1/2 * 1/2 = 1/4

So the distribution is:

P(0) = 1/4        P(1) = 1/2      P(2) = 1/4