Question

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

Let S = event that at least an odd number is the outcome of n tosses. = success

Let F = event that no odd number turns out = failure

n =1, then in one toss, probability of an odd number turning out = 3/6 = 1/2
          P(S) = 1/2   So P(F) = 1 - 1/2 = 1/2

The tosses of die are independent. The probabilities can be multiplied.

n = 2, two tosses,  P(S) = 1 -  ( P(Failure) in first toss * P(Failure) in 2nd toss)  
                                    = 1  - 1/2 * 1/2 = 1 - 1/4 = 3/4

probability distribution is 
     P(n,X) :   = 0   for n = 0
                   =  1/2  for  n = 1
                  =  3/4  for  n = 2
                  =  $1-\frac{1}{2^n}$