Menu Close

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

Tinku Tara 0 Comments
Pin




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
$$\mathrm{If}\:\mathrm{4}\:\mathrm{dice}\:\mathrm{are}\:\mathrm{thrown}\:\mathrm{together},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers} \\ $$$$\mathrm{appearing}\:\mathrm{on}\:\mathrm{them}\:\mathrm{is}\:\mathrm{13},\:\mathrm{is} \\ $$
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 11×4!=264.The partern space is A_6 ^4 =6^4 =1296.Thus the probability we need to find equal to ((264)/(1296))=((11)/(54))≈(1/5)
$$\mathrm{The}\:\mathrm{possibilities}\:\mathrm{to}\:\mathrm{success}\:\:\mathrm{are}\:\left(\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{6}\right), \\ $$$$\left(\mathrm{1},\mathrm{2},\mathrm{5},\mathrm{5}\right),\left(\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{4}\right),\left(\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{3}\right),\left(\mathrm{2},\mathrm{3},\mathrm{6},\mathrm{2}\right), \\ $$$$\left(\mathrm{3},\mathrm{3},\mathrm{3},\mathrm{4}\right),\left(\mathrm{3},\mathrm{4},\mathrm{1},\mathrm{5}\right),\left(\mathrm{3},\mathrm{4},\mathrm{2},\mathrm{4}\right),\left(\mathrm{4},\mathrm{4},\mathrm{1},\mathrm{4}\right) \\ $$$$,\left(\mathrm{2},\mathrm{2},\mathrm{3},\mathrm{6}\right),\left(\mathrm{2},\mathrm{2},\mathrm{4},\mathrm{5}\right)\:\mathrm{and}\:\mathrm{all}\:\mathrm{the}\: \\ $$$$\mathrm{permutations}\:\mathrm{of}\:\mathrm{every}\:\mathrm{set}\:\mathrm{of}\:\mathrm{four}\:\mathrm{numbers}.\mathrm{Hence}, \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{successful}\:\mathrm{cases}\:\mathrm{are} \\ $$$$\mathrm{11}×\mathrm{4}!=\mathrm{264}.\mathrm{The}\:\mathrm{partern}\:\mathrm{space}\:\mathrm{is}\: \\ $$$$\mathrm{A}_{\mathrm{6}} ^{\mathrm{4}} =\mathrm{6}^{\mathrm{4}} =\mathrm{1296}.\mathrm{Thus}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{we}\: \\ $$$$\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{264}}{\mathrm{1296}}=\frac{\mathrm{11}}{\mathrm{54}}\approx\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by bemath last updated on 19/Jun/20
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
Commented by bemath last updated on 19/Jun/20
when (2,3,4,4) the permutations is ((4!)/(2!)) = 12.
$$\mathrm{when}\:\left(\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{4}\right)\:\mathrm{the}\:\mathrm{permutations} \\ $$$$\mathrm{is}\:\frac{\mathrm{4}!}{\mathrm{2}!}\:=\:\mathrm{12}. \\ $$
Commented by bemath last updated on 19/Jun/20
i think not correct answer
$$\mathrm{i}\:\mathrm{think}\:\mathrm{not}\:\mathrm{correct}\:\mathrm{answer} \\ $$
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.((4!)/(2!))=72 The set of four numbers (3,3,3,4), (4,4,4,1)give 2.((4!)/(3!))=8possibilities.Hence ,all have 48+72+8=128 successful cases.Thus, the probability we need to find equal to ((128)/(1296))=(8/(81))≈(1/(10))
$$\mathrm{ok},\mathrm{thank}\:\mathrm{you}\:\mathrm{sir},\mathrm{I}\:\mathrm{shall}\:\mathrm{count}\:\mathrm{again}: \\ $$$$\mathrm{The}\:\mathrm{sets}\:\mathrm{of}\:\mathrm{four}\:\mathrm{numbers}\:\left(\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{6}\right), \\ $$$$,\left(\mathrm{3},\mathrm{4},\mathrm{1},\mathrm{5}\right),\mathrm{give}\:\mathrm{2}.\mathrm{4}!=\mathrm{48}\:\mathrm{possibilities} \\ $$$$\mathrm{The}\:\mathrm{sets}\:\mathrm{of}\:\mathrm{four}\:\mathrm{numbers}\:\left(\mathrm{1},\mathrm{2},\mathrm{5},\mathrm{5}\right), \\ $$$$\left(\mathrm{3},\mathrm{3},\mathrm{6},\mathrm{1}\right),\left(\mathrm{3},\mathrm{3},\mathrm{5},\mathrm{2}\right),\left(\mathrm{4},\mathrm{4},\mathrm{2},\mathrm{3}\right),\left(\mathrm{2},\mathrm{2},\mathrm{3},\mathrm{6}\right), \\ $$$$\left(\mathrm{2},\mathrm{2},\mathrm{4},\mathrm{5}\right)\:\mathrm{give}\:\:\mathrm{6}.\frac{\mathrm{4}!}{\mathrm{2}!}=\mathrm{72} \\ $$$$\mathrm{The}\:\mathrm{set}\:\mathrm{of}\:\mathrm{four}\:\mathrm{numbers}\:\left(\mathrm{3},\mathrm{3},\mathrm{3},\mathrm{4}\right), \\ $$$$\left(\mathrm{4},\mathrm{4},\mathrm{4},\mathrm{1}\right)\mathrm{give}\:\:\mathrm{2}.\frac{\mathrm{4}!}{\mathrm{3}!}=\mathrm{8possibilities}.\mathrm{Hence}\:,\mathrm{all}\:\mathrm{have} \\ $$$$\mathrm{48}+\mathrm{72}+\mathrm{8}=\mathrm{128}\:\mathrm{successful}\:\mathrm{cases}.\mathrm{Thus}, \\ $$$$\mathrm{the}\:\mathrm{probability}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\frac{\mathrm{128}}{\mathrm{1296}}=\frac{\mathrm{8}}{\mathrm{81}}\approx\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$ \\ $$
Commented by bemath last updated on 19/Jun/20
(1,1,6,5) not include sir?
$$\left(\mathrm{1},\mathrm{1},\mathrm{6},\mathrm{5}\right)\:\mathrm{not}\:\mathrm{include}\:\mathrm{sir}? \\ $$
Commented by bemath last updated on 19/Jun/20
i got ((136)/6^4 ) = ((17)/(162))
$$\mathrm{i}\:\mathrm{got}\:\frac{\mathrm{136}}{\mathrm{6}^{\mathrm{4}} }\:=\:\frac{\mathrm{17}}{\mathrm{162}} \\ $$
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
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir},\mathrm{you}\:\mathrm{are}\:\mathrm{right},\mathrm{I}\:\mathrm{missed}\:\mathrm{that}\:\mathrm{case} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *