poornima285
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if we roll 4 dice, how many outcomes are there that total to 9? also how many outcomes are there that comes with 1 or 2? Please help me with this questio.
rolling dice
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\(\displaystyle \left(\sum_{k=1}^{6}x^{k}\right)^{4}=\left(x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}\right)^{4}\).....[1]
Expand out and look at the coefficient of \(\displaystyle x^{9}\). That is the total number of outcomes that sum to 9.
Or, expand out \(\displaystyle x^{4}\left(1+x+x^{2}+x^{3}+x^{4}+x^{5}\right)^{4}=x^{4}\cdot\frac{1}{(1-x)^{4}}\)....[2]
The coefficient of \(\displaystyle x^{9}\) in [1] is the coefficient of \(\displaystyle x^{5}\) in [2].
Note the identity: \(\displaystyle \frac{1}{(1-x)}^{n}}=1+\binom{1+n-1}{1}x+\binom{2+n-1}{2}x^{2}+....+\binom{r+n-1}{r}x^{r}+....\)
Thus, the coefficient is \(\displaystyle \binom{5+4-1}{5}\). That is the total outcomes summing to 9 when rolling 4 dice.
Expand out and look at the coefficient of \(\displaystyle x^{9}\). That is the total number of outcomes that sum to 9.
Or, expand out \(\displaystyle x^{4}\left(1+x+x^{2}+x^{3}+x^{4}+x^{5}\right)^{4}=x^{4}\cdot\frac{1}{(1-x)^{4}}\)....[2]
The coefficient of \(\displaystyle x^{9}\) in [1] is the coefficient of \(\displaystyle x^{5}\) in [2].
Note the identity: \(\displaystyle \frac{1}{(1-x)}^{n}}=1+\binom{1+n-1}{1}x+\binom{2+n-1}{2}x^{2}+....+\binom{r+n-1}{r}x^{r}+....\)
Thus, the coefficient is \(\displaystyle \binom{5+4-1}{5}\). That is the total outcomes summing to 9 when rolling 4 dice.
poornima285
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the answer given is 40 for the number ofoutcomes that total 9 when 4 dice are rolled. but when i do (5+4-1C5) i am not getting the answer. could you please help me on this
There are 56 total outcomes, not 40, regardless of the answer given. Unless there is something going on I am not aware of.
How many come with a 1 or a 2?. There are 6^4=1296 total outcomes. Try counting how many of these have a 1 or 2 in them.
You could use the binomial probability and multiply that by 1296 to find the number of outcomes with a 1 or 2.
\(\displaystyle 1296\sum_{k=1}^{4}\binom{4}{k}(\frac{1}{3})^{k}(\frac{2}{3})^{4-k}=1296\cdot\frac{65}{81}\)
This binomial gives the probability of getting 'at least one' 1 or 'at least one' 2. That is why the probability is 1/3 instead of 1/6.
There are always more than one way to do counting problems.
How many come with a 1 or a 2?. There are 6^4=1296 total outcomes. Try counting how many of these have a 1 or 2 in them.
You could use the binomial probability and multiply that by 1296 to find the number of outcomes with a 1 or 2.
\(\displaystyle 1296\sum_{k=1}^{4}\binom{4}{k}(\frac{1}{3})^{k}(\frac{2}{3})^{4-k}=1296\cdot\frac{65}{81}\)
This binomial gives the probability of getting 'at least one' 1 or 'at least one' 2. That is why the probability is 1/3 instead of 1/6.
There are always more than one way to do counting problems.
Hello, poornima285!
You're right, Galactus . . . I blew it!
. . \(\displaystyle \begin{array}{cc}\text{Combination} & \text{Possible orders} \\ \hline (1,1,1,6) & _4P_{_3} \:=\:4\\ (1,1,2,5) & _4P_{_2} \:=\:12 \\ (1,1,3,4) & _4P_{_2} \:=\: 12 \\ (1,2,2,4) & _4P_{_2} \:=\:12 \\ (1,2,3,3) & _4P{_{_2} \:=\:12 \\ (2,2,2,3) & _4P_{_3} \:=\:4\end{array}\)
. . . . . . .\(\displaystyle \uparrow\)
. . Missed this one!
\(\displaystyle \text{There are: }\:4+12+12+12+12 + 4\:=\:\boxed{56}\text{ outcomes that total 9.}\)
I assume this mean at least one 1 or one 2 . . .
\(\displaystyle \text{There are: }\:6^4 \:=\:1296\text{ possible outcomes.}\)
\(\displaystyle \text{There are: }\,4^4 \:=\:256\text{ outcomes with }no\text{ 1 or 2.}\)
\(\displaystyle \text{Therefore, there are: }\:1296 - 256 \:=\:1040\text{ outcomes with a 1 or a 2.}\)
You're right, Galactus . . . I blew it!
. . \(\displaystyle \begin{array}{cc}\text{Combination} & \text{Possible orders} \\ \hline (1,1,1,6) & _4P_{_3} \:=\:4\\ (1,1,2,5) & _4P_{_2} \:=\:12 \\ (1,1,3,4) & _4P_{_2} \:=\: 12 \\ (1,2,2,4) & _4P_{_2} \:=\:12 \\ (1,2,3,3) & _4P{_{_2} \:=\:12 \\ (2,2,2,3) & _4P_{_3} \:=\:4\end{array}\)
. . . . . . .\(\displaystyle \uparrow\)
. . Missed this one!
\(\displaystyle \text{There are: }\:4+12+12+12+12 + 4\:=\:\boxed{56}\text{ outcomes that total 9.}\)
I assume this mean at least one 1 or one 2 . . .
\(\displaystyle \text{There are: }\:6^4 \:=\:1296\text{ possible outcomes.}\)
\(\displaystyle \text{There are: }\,4^4 \:=\:256\text{ outcomes with }no\text{ 1 or 2.}\)
\(\displaystyle \text{Therefore, there are: }\:1296 - 256 \:=\:1040\text{ outcomes with a 1 or a 2.}\)
There it is 
You hardly blew it. An easy mistake. Counting problems are very easy to do that with. I miscounted one the other day until pka pointed out my error.
Your explanation for the other half of the problem is easier than mine. When dealing with 'at least one' problems it is most always best to use the complement.
You hardly blew it. An easy mistake. Counting problems are very easy to do that with. I miscounted one the other day until pka pointed out my error.
Your explanation for the other half of the problem is easier than mine. When dealing with 'at least one' problems it is most always best to use the complement.
poornima285
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thanks a lot for your replies. they were very helpful.