Why am I getting probability greater than 1 for rolling a 4 twice?

[huan.conchito]

Roll a fair die twice and find the probability of at least one 4
here's what i did:

P(A) = |A|/|Ω| = (6C1*6C0+6C1*6C1)/(6C1*6C1)
And i get the answer as 42/36 and the real answer is 5/36
what did I do wrong

 

[xanthym]

Roll a fair die twice and find the probability of at least one 4
here's what i did:

P(A) = |A|/|Ω| = (6C1*6C0+6C1*6C1)/(6C1*6C1)
And i get the answer as 42/36 and the real answer is 5/36
what did I do wrong

Although enumeration isn't the most efficient solution to probability problems, sometimes it can show where errors are occurring. All possible 36 outcomes of 2 rolls of a fair die are shown below. Those containing At Least One (1) "Four (4)" are highlited. By counting the highlited cases below, it can be seen that exactly (11) cases out of (36) show at least one (1) "Four (4)". Therefore, the required probability is (11/36).

Roll
AB
==
11
12
13
14
15
16

21
22
23
24
25
26

31
32
33
34
35
36

41
42
43
44
45
46

51
52
53
54
55
56

61
62
63
64
65
66
==

~~

 

[jdavel]

huan,

xanthym is right; it's hard to argue with a 100% coverage test!

Another way to get the same answer (and it works on LOTS of probability problems) relies on the fact that "getting at least one 4" and "getting exactly zero 4s" are mutually exclusive and collectively exhaustive. In other words, all possible outcomes are included in one or the other but not both of these. So their probabilities have to sum to 1. But the probability of "getting exactly zero 4s" is easy to calculate; it's just 5/6 *5/6 = 25/36. So the prob of "getting at least one 4" is 1- 25/36 = 11/36