A sum between 5 and 10

Video Transcript

So here for this question we suppose the same of the two dies as as supposes it? S and we know that for two dies And the same could be between two and 12, Smallest is two and largest as 12. And it is easy to figure out that one assets between two and 7 and we suppose we divide them into die one and die. So we could have s um one plus s minus one or two plus s minus ta and blah blah blah As -1- one. Right? So well as as equal to two we just have one possibility whereas as equal to three we have two possibility as is equal to seven. We have six possibilities. Six possibilities. Right? So in each case we have as -1 possibilities And well as as equal to eight We do not have 17 because The largest value probably six. So we have 26 until 62. So we have five Let's say we have five Possibilities here for us equal to nine. We have four possibilities. Similarly for 10. We have three for 11. We have two for 12. We have one possibility and then we can calculate the probability in each case. And then the 1st 1 a no no more than four. So as more than or equal to four we have as equal to two, so as equal to 234. In total we have one plus two plus three possibilities. And for the combination of two dies We should have six times six. Right, so this property is equal to um six over 66 times six is to be equal to 1/6 years. For b As not less than eight. So grades are equal to eight. So we just need to sum up this this this this and this value. No. S could be 89, 10, 11 and 12. So that probably is equal to five plus four plus four plus three plus two plus one over 36. And then this is a quarter um 15 out of 36 which is equal to 5/12. This is the second property For the last one and between five and 10 exclusive. So this means that as could be equal to As is equal to 6, 7, 8 and nine. So probability is for six the possibility is 5476, 48. We have 549, we have four out of 36. And then this problem is equal to um 20 so 20 over 36. So we should have five out of nine which is the last probability here. So the second probability the first probability right.

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Question 1140156:
A pair of dice is rolled. Find the probability of rolling
​a) a sum not more than 4​,
​b) a sum not less than 10​,
​c) a sum between 3 and 7 ​(exclusive).
i finished part A and B but how would you solve part C) please and thank you
Answer by AnlytcPhil(1761)   (Show Source):

You can put this solution on YOUR website!
a) a sum not more than 4​,

That's the 6 rolls colored red below out of the 36 possible rolls, a
probability of 6 out of 36 or a probability of 6/36 which reduces to
1/6. 

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

​

b) a sum not less than 10​,

That's the 6 rolls colored red below out of the 36 possible rolls, a
probability of 6 out of 36 or a probability of 6/36 which reduces to
1/6.

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

​

c) a sum between 3 and 7 ​(exclusive).

To exclude 3 and 7 and be between them means to use 4,5, and 6 only.
That's the 12 rolls colored red below out of the 36 possible rolls, a
probability of 12 out of 36 or a probability of 12/36 which reduces to
1/3.  

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

Edwin

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Question 915392: My sum is between 5 and 10. I am a double plus one. Who could I be?
Answer by KMST(5315)   (Show Source):

You can put this solution on YOUR website!
Are we talking about the sum of the digits of a number?
Is that a 2-digit number? A 3-digit number?

IF we are talking about a 2-digit number,
whose sum of digits is between 5 and 10 (including 5 and 10),
and the number is a double plus 1 (meaning an odd number),
there are many possibilities.
Among odd 2-digit numbers ending in 1,
the numbers whose sum of digits is 5, 6, 7, 8, 9, or 10 are:
41, 51, 61, 71, 81, and 91.
Among odd 2-digit numbers ending in 3,
the numbers whose sum of digits is 5, 6, 7, 8, 9, or 10 are:
23, 43, 53, 63, and 73.
Among odd 2-digit numbers ending in 5,
the numbers whose sum of digits is 6, 7, 8, 9, or 10 are:
15, 25, 35, 45, and 55 (and there is none with 2 digits adding to 5).
Among odd 2-digit numbers ending in 7,
the numbers whose sum of digits is 8, 9, or 10 are:
17, 27, and 37.
Finally, among odd 2-digit numbers ending in 9,
the only numbers whose sum of digits is between 5 and 10 is
19.

What is the sum between 1 and 10?

What is the probability that the sum is 5?

What is the probability for a pair of dice to show a sum of 5 or 10?