If-4-dice-are-thrown-together-then-the-probability-that-the-sum-of-the-numbers-appearing-on-them-is-13-is-

Question Number 99142 by babar last updated on 18/Jun/20

If 4 dice are thrown together, then the

probability that the sum of the numbers

appearing on them is 13, is

Answered by 1549442205 last updated on 19/Jun/20

The possibilities to success are (1, 2, 4, 6),

(1, 2, 5, 5), (2, 3, 4, 4), (2, 3, 5, 3), (2, 3, 6, 2),

(3, 3, 3, 4), (3, 4, 1, 5), (3, 4, 2, 4), (4, 4, 1, 4)

, (2, 2, 3, 6), (2, 2, 4, 5) and all the

permutations of every set of four numbers . Hence,

the sum of the successful cases are

11 \times 4! = 264 . The partern space is

A_{6}^{4} = 6^{4} = 1296 . Thus the probability we

need to find equal to \frac{264}{1296} = \frac{11}{54} \approx \frac{1}{5}

Commented by bemath last updated on 19/Jun/20

great sir

Commented by bemath last updated on 19/Jun/20

when (2, 3, 4, 4) the permutations

is \frac{4!}{2!} = 12 .

Commented by bemath last updated on 19/Jun/20

i think not correct answer

Commented by 1549442205 last updated on 19/Jun/20

ok, thank you sir, I shall count again :

The sets of four numbers (1, 2, 4, 6),

, (3, 4, 1, 5), give 2.4! = 48 possibilities

The sets of four numbers (1, 2, 5, 5),

(3, 3, 6, 1), (3, 3, 5, 2), (4, 4, 2, 3), (2, 2, 3, 6),

(2, 2, 4, 5) give 6 . \frac{4!}{2!} = 72

The set of four numbers (3, 3, 3, 4),

(4, 4, 4, 1) give 2 . \frac{4!}{3!} = 8 possibilities . Hence, all have

48 + 72 + 8 = 128 successful cases . Thus,

the probability we need to find equal to

\frac{128}{1296} = \frac{8}{81} \approx \frac{1}{10}

Commented by bemath last updated on 19/Jun/20

(1, 1, 6, 5) not include sir ?

Commented by bemath last updated on 19/Jun/20

i got \frac{136}{6^{4}} = \frac{17}{162}

Commented by bemath last updated on 19/Jun/20

Commented by 1549442205 last updated on 19/Jun/20

Thank you sir, you are right, I missed that case