top button
Flag Notify
    Connect to us
      Site Registration

Site Registration

Two digits are selected at random from digits 1 through 9. If sum is even, find the probability of both numbers are odd?

0 votes
840 views

Two digits are selected at random from digits 1 through 9. If sum is even, find the probability that both numbers are odd?

posted Nov 10, 2017 by anonymous

Share this puzzle
Facebook Share Button Twitter Share Button LinkedIn Share Button

1 Answer

0 votes

Out of the digits 1 through 9, 5 are odd and 4 are even.

The sum of the two digits is odd precisely when one digit is odd and the other digit is even, so
P(sum is odd) = P(odd, then even) + P(even, then odd).

It is not clear if these two digits are selected with or without replacement, so I will use both interpretations.

Without replacement, P(odd sum) = (5/9)(4/8) + (4/9)(5/8) = 5/9.
With replacement, P(odd sum) = (5/9)(4/9) + (4/9)(5/9) = 40/81.

answer Dec 28, 2017 by Yasin Hossain Siinan
Andha he kya ??ques thik se par
...