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  • Let's say I have 2 die
  • I roll them

What is the probability that value on die 1 + value on die 2 is greater than 4 ?

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    You have $36$ possible outcomes: write all them on a piece of paper and count those summing to a number $> 4$. It will give you a probability of $31/36$. – 2017-01-08
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    @aduh Do you assume the roles are independent? - Yes Are the die weighted? - no What's your probability model? - Didn't get you – 2017-01-08
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    @crostul Let's say I have 10 dice, This solution is not feasible – 2017-01-08

2 Answers 2

1

Set $S=\{1,2,\cdots,6\}$ and define $A=\{(x,y)\in S^2 |\,x+y> 4\}$. We have $$\mathbb{P}(A^c)=\frac{6}{36}$$

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    Sir care you to explain Mathematical noob here – 2017-01-08
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    $$A^c=\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\}$$ and the sample space is as follow $$\{(1,1),(1,2),\cdots,(1,6),\cdots,(6,1)\cdots,(6,6)\}$$ – 2017-01-08
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    But sir 6/36 is not the correct solution – 2017-01-08
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    $$P(A)=1-P(A^c)=1-\frac {6}{36}=\frac{30}{36}=\frac 56$$ – 2017-01-08
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    Thank you sir I owe you with my life – 2017-01-08
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    Good luck. -:). – 2017-01-08
2

There are 36 possible outcomes in total. One possibility to have a sum of 2 (1+1), 2 possibilities to have a sum of 3 (2+1, 1+2) and 3 possibilities to have a sum of 3 (1+3,3+1,2+2). Hence, six out of 36 possible outcomes are less than or equal to 4. We obtain

$$P(\text{sum} > 4) = 1 - \frac{6}{36}=\frac{30}{36}=\frac{5}{6}$$

This is all under the assumption that the dice are fair.