Question

If 3 dice are thrown, what is the probability that the number appeared on one dice is equal to the sum of the numbers appeared on the other two dice?

Answer 1

@ganeshie8

Answer 2

@wolf1728

Answer 3

If 3 dice are thrown, what is the probability that the number appeared on one dice is equal to the sum of the numbers appeared on the other two dice?

Answer 4

2,(1,1) upto 6 (1,5) (3,3)(2,4),(4,2)(5,1)so total=15

Answer 5

n(s)=6^3=216,so=15/216

Answer 6

this looks hard

Answer 7

me too

Answer 8

the probab q

Answer 9

what is HCF ?

Answer 10

probability question

Answer 11

I have edited the problem, check it now @ikram002p

Answer 12

If 3 dice are thrown, what is the probability that the number appeared on one dice is equal to the sum of the numbers appeared on the other two dice?

*confused *

Answer 13

if the number appeared in one dice is 2 then other two will be 2 i.e (sum of two dice is2) (1,1)
if 3 (1,2) (2,1)

Answer 14

ok so we have 3 dice
mmm

1,1 , 2
1,2 , 3
1,3 ,4
1,4,5
1,5,6

2,1 ,3
2,2,4
2,3,5
2,4,6

3,1,4
3,2,5
3,3,6

4,1,5
4,2,6

5,1,6

Answer 15

does order of the other two dice important ?

Answer 16

no

Answer 17

ok
so grouping
(1,2,3,4,5) (1,2,3,4,5) =(1,1) (1,2) (1,3) (1,4) (1,5) (2,3) (2,4) (3,3)
mmmm

Answer 18

P (of sum ) = p( of having sum 2 ) + p( of having sum 3) + p( of having sum 4) + P (of having sum 5) + p( of having sum 6)

Answer 19

p (of having sum 2) is 1/6 *1/6* 1/6
only when first dice =1 and second =1 and third = 2

Answer 20

p( of having sum 3 ) is 2/6 * 2/6 *1/6

(1,2,3) or (2,1,3)

Answer 21

@ganeshie8 mmm idk a bit confused

Answer 22

Four-digit numbers are formed using the digits 0, 2, 3, 4, 5 without repetition. What is the probability that the numbers are divisible by 4?

Answer 23

04,12,24,32,52 are last two numbers which is divisible by 4
so 5*3*2=30

Answer 24

what we are solving lol !!!

Answer 25

ic , so is it other Qn ?

Answer 26

ok i would rather to find whole space :P

Answer 27

ya.i have lot of question to solve.its so much time for one question so atleast i can get answer to one question.its so boring if not getting answer

Answer 28

ok ok for first one i think finding the whole space which is 15/6^3

Answer 29

If 3 dice are thrown, what is the probability that the number appeared on one dice is equal to the sum of the numbers appeared on the other two dice?


answer 15/6^3

Answer 30

no.that's wrong .

Answer 31

:O
why
whats the right ansewer ?

Answer 32

i don't kno

Answer 33

could it be like this ?
(1/6)^3 + (2/6) (1/6)^2+ (3/6)(2/6)(1/6)

Answer 34

15/6^3
if we ignore order
but
9/6^3
for order

Answer 35

i know options 1/24,5/216,5/24,5/36,something like that

Answer 36

ohh so
9/6^3=1/24

mmm xD
* confused*

Answer 37

four digit number

Answer 38

correct answer is
http://www.wolframalpha.com/input/?i=45%2F216

Answer 39

i have just listed down all the possibilities, not sure if there is a clever way.. .

Answer 40

mm for which Qn ?

Answer 41

for the original main question,
@dg2 kindly limit to one question per post so that there won't be any confusions :)

Answer 42

huh yeah lol its confusing me
so ganesh with listing posibilities
u took it like this ?
(2,1,1)
(3,1,2)
(3,2,1)
(4,1,3)
(4,3,1)
(4,2,2 )
??

Answer 43

```
1 : 1 1 2
2 : 1 2 1
3 : 1 2 3
4 : 1 3 2
5 : 1 3 4
6 : 1 4 3
7 : 1 4 5
8 : 1 5 4
9 : 1 5 6
10 : 1 6 5
11 : 2 1 1
12 : 2 1 3
13 : 2 2 4
14 : 2 3 1
15 : 2 3 5
16 : 2 4 2
17 : 2 4 6
18 : 2 5 3
19 : 2 6 4
20 : 3 1 2
21 : 3 1 4
22 : 3 2 1
23 : 3 2 5
24 : 3 3 6
25 : 3 4 1
26 : 3 5 2
27 : 3 6 3
28 : 4 1 3
29 : 4 1 5
30 : 4 2 2
31 : 4 2 6
32 : 4 3 1
33 : 4 5 1
34 : 4 6 2
35 : 5 1 4
36 : 5 1 6
37 : 5 2 3
38 : 5 3 2
39 : 5 4 1
40 : 5 6 1
41 : 6 1 5
42 : 6 2 4
43 : 6 3 3
44 : 6 4 2
45 : 6 5 1
```

Answer 44

yes @ganeshie8

Answer 45

if that helps in reverse engineering... @ikram002p

Answer 46

@dg2 - i suggest you move on with your other problems, we will tag you once we figure out some solution :)

Answer 47

well i believe the equation is asking about the sum of 2 numbers

Answer 48

no the difference combined

Answer 49

shall i close n post a new question

Answer 50

sure :)

Answer 51

Solve in in integers :

1<=(a,b,c)<=6

a = b+c
or
b = c+a
or
c = a+b

Answer 52

have nothing else to say @ganeshie8 lol
the order thing confuse me
i only thought
first dice is the one which should be the sum
so just multiply pre approch with 5
5 ( 1/6^3+2/6 1/6^2+ 3/6 2/6 1/6 )

Answer 53

thats working xD
how did u get that ? care to explain it step by step

Answer 54

http://www.wolframalpha.com/input/?i=5%281%2F6%5E3%2B2%2F6*1%2F6%5E2%2B+3%2F6+*2%2F6+*1%2F6%29

Answer 55

i know
will possibilty without ordering is
1/6^3+2/6 1/6^2+ 3/6 2/6 1/6

so we have 5 orderd since only sum we can get 2,3,4,5,6

Answer 56

how this works :
1/6^3+2/6 1/6^2+ 3/6 2/6 1/6

?

Answer 57

break it down please, my brain is malfunctioning today >.<

Answer 58

It cannot be the case that more than one die is equal to the sum of the other two. So there is either 1 die that is the sum of the other two or 0 dice have that property.

If there is 1 die, it must be the case that it is the one showing the greatest value. That value must be at least 1 more than the next largest die (or dice in case of a tie for second place).

Answer 59

look pre comments \(\uparrow\)
p= p( of having sum 2 ) + p( of having sum 3) + p( of having sum 4) + P (of having sum 5) + p( of having sum 6)
= 1/6^3+2/6 1/6^2+ 3/6 2/6 1/6

huh its something i tool in Abstract algebra about order grouping
is we have odd set like (2,3,4,5,6) ( set of sum ) its of order 5 so directly



in propabilty we can iterept it like this :-
p( of having sum 2 ) = 1/6 * 1/6 * 1/6
p( of having sum 3) =1/6* 1/6*1/6
p( of having sum 4) =2/6 *1/6*1/6
p(of having sum 5) = 2/6*1/6*1/6
p(of having sum 6) =3/6*1/6*1/6

Answer 60

So all three dice tying is out. The chance of getting a number on one die is 1. The chance of matching that number on the second die is 1/6 and of matching on the other die is 1/6. So the chance of hitting a triple is 1 x 1/6 x 1/6 = 1/36 so put 1/36 in the NOT category.

Answer 61

got that Lars, only one die equals the sum of other two. This follows directly from below :

if a = b+c, then b = a-c, c = a-b

Answer 62

ok how many possible combinations are there?

Answer 63

Oh that brilliant! so you want to work out the complement first is it

Answer 64

yes

Answer 65

u can get
sum of 2 form 1 compination
sum of 3 from 1 compination
sum of 4 from 2 compination
sum of 5 from 2 compination
sum of 6 from 3 compination

Answer 66

I'm not liking this tak. How about the sum of the dice must be an even number?

Answer 67

it says sum of two dices

Answer 68

If the number on one die equals the sum on the other two dice, then the sum on all three dice must be even.

Answer 69

If the greatest die is odd, then there must be exactly one other odd die.

Answer 70

Case of greatest die is odd:

Die is 5, exactly one of the other dice must be 3 or 1
Die is 3, exactly one of the other dice must be 1

Answer 71

If the greatest die is even: both the other dice must be even or both must be odd.

Answer 72

beautiful !!

Answer 73

Let's back up a little bit. I will make the bold claim that the chance that the sum of three dice summing to an odd number is .5. Can we prove that?

Answer 74

odd cases :
5 : (3, 2) , (1, 4) : 12 ways
3 : (1, 2) : 6 ways

even cases :
6 : (1, 5), (3, 3), (2, 4), : 15 ways
4 : (1, 3), (2, 2) : 9 ways
2 : (1, 1) : 3 ways

------------------------------
45 ways

this looks to be the shortest method possible xD

Answer 75

wow !

Answer 76

we can prove that claim easily - the sum of 3 numbers is either odd or even, by symmetry we can conclude that the chances of getting sum odd is 1/2

Answer 77

this is how it works in odd order grouping xD

Answer 78

i see

Answer 79

Okay, I see the proof is more or less trivial.

The chances of the total (of three) being odd is .5 which is OUT
Now we figured the chance of all being tied at 1/36 but half of those are odd totals already excluded. So three tied and even = 1/72 which is OUT.

Answer 80

Humm running out of gas here. I'll look back and see how things are going the other way.

Answer 81

216 is the total cases. I eliminated half of them leaving 108. Then I eliminated another 1/72 of them, leaving 105.

@ganeshie8 accounts for 45 ways of winning. So I am looking for a genius way of eliminating 60 cases.

Answer 82

108 - odd
3 - even and all dice equal
----------------
111 ways are eliminated so far, leaving 105 ways

Answer 83

Random hypotheses:

If exactly two dice are tied -> neither can be the greatest die and the greatest must be even.

Answer 84

yup! two dice can be tied in 6 ways

Answer 85

oh wait, the converse is not true

Answer 86

It cannot be true that the top die is 6 and only one of the others is odd.
It cannot be true that the top die is 5 and both of the others is even.
It cannot be true that the top die is 4 and only one of the others is odd.
It cannot be true that the top die is 3 and both of the others is even.
It cannot be true that the top die is 2 and only one of the others is odd.

The case that the top die is 1 was eliminated by excluding odd totals of all three dice.

Answer 87

Hmm...bored to death, and the drunk woke up and is talking to space aliens.