Probability question ?

Hi,

In your questions there are total 05 occurrences and each occurrence have 02 results for the first team ie, win or lose for a. Lets call these results W or L. Now there are following possibilities for the first teams

1) All Win => 05 W and the combinations would be 5c5 = 5!/[5!*(5-5)!] = 5!/[5!*0!] = 120/[120*1] = 1

2) 04 Win and 01 lose => 04 W and the combinations would be 5c4 = 5!/[4!*(5-4)!] = 5!/[4!*1!] = 120/[24*1] = 5

3) 03 Win and 02 lose => 03 W and the combinations would be 5c3 = 5!/[3!*(5-3)!] = 5!/[3!*2!] = 120/[6*2] = 10

4) 02 Win and 03 lose => 02 W and the combinations would be 5c2 = 5!/[2!*(5-2)!] = 5!/[2!*3!] = 120/[2*6] = 10

5) 01 Win and 04 lose => 01 W and the combinations would be 5c1 = 5!/[1!*(5-1)!] = 5!/[1!*4!] = 120/[1*24] = 5

6) 00 Win and 05 lose => 00 W and the combinations would be 5c0 = 5!/[0!*(5-0)!] = 5!/[0!*5!] = 120/[1*120] = 1

Therefore total combinations will be = sum of all above = 1 + 5 + 10 + 10 + 5 + 1 = 32

This can also be done by: 2^5 = 32 but I personally like the above mentioned approach as it is more clear in concept.

Hope I helped. Bye.

12

Not sure about your question but it seems there are 6 games each of which can have two possible outcomes.

The total of the possible outcomes for the six games is 2^6, = 64.