0

Here is a question. Two fair dice are thrown, find the probability sum of outcomes is 11.

case 1: considering the dice to be distinct, (6,5) and (5,6) are the favourable outcomes out of total of 6x6=36 outcomes. P(E) = 2/36 = 1/18

Case 2: considering the dice to be identical, only 1 favourable outcome is (5,6). For total number of cases, 1 can link with 6 elements (1 to 6), however, 2 can now link with only 5 elements (2 to 6) as (1,2) has already been counted and as dice are identical (1,2) is same as (2,1). Hence, total number of cases are 6x7/2 = 21 Therefore, P(E) = 1/21 Can someone help me understand what is really going on here.

Mitansh
  • 31
  • 1
    Do the answers to this older question help? https://math.stackexchange.com/questions/2240392/rolling-2-dice-not-using-36-as-the-base/ – Especially Lime May 16 '24 at 09:13
  • As the first comment by @lulu there says, the problem is that the $21$ cases are not equiprobable. ${1,2}$ is twice as likely as ${1,1}$. – Henry May 16 '24 at 09:23
  • Yes, the answers to the older question definitely helped. My apologies for not going over that before asking my question. – Mitansh May 16 '24 at 09:27

0 Answers0