Die probability

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Zelda22

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A decimal die is tossed three times in order to generate a sequence of three random numbers.


a. Find the probability that the sequence obtained has its numbers in increasing order from left to right; for example, 5-7-8.
b. Find the probability that the sequence obtained has its second number larger than the other two; for example, 3-6-2.

I'm not sure how to solve this.

I tried to write down the numbers that meet those requirements. I know that a die tossed three times has 216 possible outcomes.

a. I got 15 combinations, would the answer should be 15/216 ???
b. I got 55 combinations, would the answer should be 55/216 ???

Can someone explain to me how to solve it, please? Thanks
 
Zelda22 said:
A decimal die is tossed three times in order to generate a sequence of three random numbers.


a. Find the probability that the sequence obtained has its numbers in increasing order from left to right; for example, 5-7-8.
b. Find the probability that the sequence obtained has its second number larger than the other two; for example, 3-6-2.

I'm not sure how to solve this.

I tried to write down the numbers that meet those requirements. I know that a die tossed three times has 216 possible outcomes.

a. I got 15 combinations, would the answer should be 15/216 ???
b. I got 55 combinations, would the answer should be 55/216 ???

Can someone explain to me how to solve it, please? Thanks
Click to expand...
How many faces (with unique numbers) does a decimal die have?

Please list you 15 combinations for part 'a'?
 
Subhotosh Khan said:
How many faces (with unique numbers) does a decimal die have?

Please list you 15 combinations for part 'a'?
Click to expand...
oh, I realized now, it's a decimal die, not a regular die, I need to redo all my numbers, oh boy!, but once I get my answers, should be the numbers of combinations over 1000?

Zelda22 said:
oh, I realized now, it's a decimal die, not a regular die, I need to redo all my numbers, oh boy!, but once I get my answers, should be the numbers of combinations over 1000?
Click to expand...
This particular decimal die has number ( 0,1,2,3,4,5,6,7,8,9) no sure if will made any difference.
 
Zelda22 said:
oh, I realized now, it's a decimal die, no a regular die, I need to redo all my numbers, oh boy!, but once I get my answers, should be the numbers of combinations over 1000?
Click to expand...
Hi Zelda,
Does the order of the outcome matter? For example, if I get 8, 5, 7. I can rearrange the number to get increasing orders 5,7,8. Or does it have to come out exactly as 5, 7,8 (no rearrangement allowed)?
:)
 
AvgStudent said:
Hi Zelda,
Does the order of the outcome matter? For example, if I get 8, 5, 7. I can rearrange the number to get increasing orders 5,7,8. Or does it have to come out exactly as 5, 7,8 (no rearrangement allowed)?
:)
Click to expand...
I'm not sure, I think it should be the order you got them.
 
Zelda22 said:
A decimal die is tossed three times in order to generate a sequence of three random numbers.


a. Find the probability that the sequence obtained has its numbers in increasing order from left to right; for example, 5-7-8.
b. Find the probability that the sequence obtained has its second number larger than the other two; for example, 3-6-2.

I'm not sure how to solve this.

I tried to write down the numbers that meet those requirements. I know that a die tossed three times has 216 possible outcomes.

a. I got 15 combinations, would the answer should be 15/216 ???
b. I got 55 combinations, would the answer should be 55/216 ???

Can someone explain to me how to solve it, please? Thanks
Click to expand...
Have you learned counting with factorial? More specifically, combination and permutation?
 
BigBeachBanana said:
Have you learned counting with factorial? More specifically, combination and permutation?
Click to expand...
No.
This particular decimal die has number ( 0,1,2,3,4,5,6,7,8,9) no sure if will made any difference.

I was thinking for b. (9x9)+(8x8)+(7x7)+(6x6)+(5x5)+(4x4)+(3x3)+(2x2)+(1x1)=285 , then It would be 285/1000 = 57/200 ???
 
Zelda22 said:
This particular decimal die has number ( 0,1,2,3,4,5,6,7,8,9) no sure if will made any difference.
Click to expand...
No, as long as they're unique it doesn't make a difference.
I got an answer by using combination/ permutation. But if you haven't learned them yet, I don't see an efficient solution besides listing them out which is toilsome.

I haven't looked at b) yet.
 
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Zelda22 said:
A decimal die is tossed three times in order to generate a sequence of three random numbers.


a. Find the probability that the sequence obtained has its numbers in increasing order from left to right; for example, 5-7-8.
b. Find the probability that the sequence obtained has its second number larger than the other two; for example, 3-6-2.
Click to expand...
But if so then there are [imath]10^3=1000[/imath] possible triples.
Assuming that increasing means that each term is greater than the preceeding.
EX: [imath](0,4,7)[/imath] and not [imath](1,1,9)[/imath]
I get [imath]36[/imath] triples that begin [imath](0,X,Y)[/imath] where [imath]0<X<Y\le 9[/imath].
Can you explain that?
[imath][/imath]
 
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well, I'm not sure but starting with the first digit
(0) 8+7+6+5+4+3+2+1 combinations
(1) 7+6+5+4+3+2+1
(2) 6+5+4+3+2+1
(3) 5+4+3+2+1
(4) 4+3+2+1
(5) 3+2+1
(6) 2+1
(7) 1

I'm not sure if that is correct
 
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Zelda22 said:
well, I'm not sure but starting with the first digit
(0) 8+7+6+5+4+3+2+1 combinations
(1) 7+6+5+4+3+2+1
(2) 6+5+4+3+2+1
(3) 5+4+3+2+1
(4) 4+3+2+1
(5) 3+2+1
(6) 2+1
(7) 1

I'm not sure if that is correct
Click to expand...
The sum matches my answer. [imath]{10 \choose 3}=120[/imath]
 
I got my answers, but I don't know if are correct, (I wrote all the numbers, I don't know if there is a formula that I could use)
A decimal die is tossed three times in order to generate a sequence of three random numbers.


a. Find the probability that the sequence obtained has its numbers in increasing order from left to right; for example, 5-7-8.

A/ 120/1000=3/25

b. Find the probability that the sequence obtained has its second number larger than the other two; for example, 3-6-2.

A/ 285/1000=57/200

Can someone verify my answers, please? Thanks
 
Zelda22 said:
I got my answers, but I don't know if are correct, (I wrote all the numbers, I don't know if there is a formula that I could use)
A decimal die is tossed three times in order to generate a sequence of three random numbers.


a. Find the probability that the sequence obtained has its numbers in increasing order from left to right; for example, 5-7-8.

A/ 120/1000=3/25

b. Find the probability that the sequence obtained has its second number larger than the other two; for example, 3-6-2.

A/ 285/1000=57/200

Can someone verify my answers, please? Thanks
Click to expand...
Your a) matches mine, but b) does not.
 
Question: in part (b) is it still required that the three numbers be distinct?
 
Zelda22 said:
I got my answers, but I don't know if are correct, (I wrote all the numbers, I don't know if there is a formula that I could use)
A decimal die is tossed three times in order to generate a sequence of three random numbers.


a. Find the probability that the sequence obtained has its numbers in increasing order from left to right; for example, 5-7-8.

A/ 120/1000=3/25

b. Find the probability that the sequence obtained has its second number larger than the other two; for example, 3-6-2.

A/ 285/1000=57/200

Can someone verify my answers, please? Thanks
Click to expand...
For b), assuming the 3-digits are distinct, I got
[math]\frac{{2 \choose 1}{10 \choose 3}}{10^3}=\frac{240}{1000}=0.24[/math]
 
Zelda22 said:
It doesn't say that. So I assumed they aren't.
Click to expand...
The question does say, "the other two numbers". which means they are distinct otherwise it is not two.
In that case I agree that the answer is [imath]240[/imath] because in any three digits one is greater that the other two. That is the middle one.
[imath]\dbinom{10}{3}=120[/imath] which we double for symmetry.
Now however, if we are allowed to count [imath](0,X,0)~,~0<X\le9[/imath]
We can use all of the [imath]240[/imath]; but must add [imath]45[/imath] more to it.
 
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