How do you calculate: "If you roll two fair six-sided dice, what is the probability that the sum is 5 or lower?" Without having to draw a large table like so: img
Does an equation of some sort exist?
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One way or another, your task amounts to counting. To me, the table is by far the clearest way to do this, especially when you're first learning the material. The answers below have outlined some other ways to streamline the counting, but it all comes down to counting. The answer to the question "Does an equation of some sort exist?" is no -- at least, not an equation that's worth knowing. – Aaron Montgomery Oct 10 '18 at 18:47
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I agree with @AaronMontgomery. There is a general formula for this, but for a small example such as this one I think it’s more instructive to work through it yourself. – amd Oct 10 '18 at 20:27
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Without constructing a table, we can define a random variable $X$: sum of values of two dices. Then we are looking for $$P(X \le 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$$ $$= \frac{1}{36}+\frac{2}{36}+\frac{3}{36}+\frac{4}{36} = \frac{5}{18}$$
because $P(X = 2)$ has one possible combination: $(1,1)$;
$P(X = 3)$ has two: $(1,2)$ or $(2,1)$;
$P(X = 4)$ has three: $(1,3)$, $(3,1)$ or $(2,2)$;
$P(X = 5)$ has four $(1,4)$, $(4,1)$, $(2,3)$ or $(3,2)$.
ArsenBerk
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Yeah, just expect for the minor difference that a variable is random, while a process is stochastic. source. But in our case, can you refer to this as a process? – Ryan Cameron Oct 11 '18 at 22:08
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The answer is the sum of natural numbers form $1$ to $4$, which is $4(5)/2 =10.$ divided by the total number of outcomes which is $36$
That makes it $\frac {10}{36}=0.2777$
Note that if the first die is $1$, we have $4$ possible outcome for the second one. Similarly for the first one being $2$ we have $3$ possibilities for the second one and so forth.
Mohammad Riazi-Kermani
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