Question

A box contains 11 tickets numbered from 1 to 11. Two tickets are drawn at random with replacement. If the sum is even, find the probability that both the numbers are odd.

Answer 1

To get an odd number as the sum, you need to take out an odd number of balls wiht an odd number on them.
In total, there are 5 even balls and 6 odd balls.
Possible combinations are
E O , E is even, O is odd
1 5
3 3
5 1

First case: 1 even ball and 5 odd balls
=5C1×6C5
=5×6
=30

Second case: 3 even balls and 3 odd balls
= 5C3×6C3
=10×2
=200

Third case: 5 even balls and 1 odd ball
= 5C5×6C1
=1×6
=6

Total favourable combinations = 30 + 200 + 6 = 236.
Total possible combinations = 11C6 = 462

Probability = 236/462 = 118/231 = 0.5108225... approx or 51.08%

hope it helped u❤️❤️

Answer 2

❤Total tickets=11
Two tickets are drawn replacement=2

Tickets are drwan replacement
-------------------------------------------------
Total tickets

2
-----
11

5.5