Two fair dice (one red and one green) are rolled. What is the probability that the sum is 5, given that the green one is either 4 or 3?

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Two fair dice (one red and one green) are rolled. What is the probability that the sum is 5, given that the green one is either 4 or 3?

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Answer 1 · 2017-04-01T00:11:25

Probability $= \frac{1}{6}$ $= 1/6$

Explanation:

Let $A$ $A$ be the event that the sum of the two dice is $5$ $5$
Let $B$ $B$ be the event that the green die is either a $3$ $3$ or $4$ $4$

Then we want $P (A ∣ B)$ $P( A | B)$ which we calculate using the conditional probability formula:

$P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$ $P( A | B) = (P(A nn B)) / (P(B))$

Consider first $P (A \cap B)$ $P(A nn B)$ which we can calculate using

$P (A \cap B) = \frac{n (A \cap B)}{n (T)}$ $P(A nn B) = (n(A nn B)) / (n(T))$

Where, $n (T)$ $n(T)$ is the total number of possible outcomes. As there are two dices then each has $6$ $6$ possible outcomes, making $n (T) = 36$ $n(T)=36$ .

And, $n (A \cap B)$ $n(A nn B)$ is the number of outcomes where the total is $5$ $5$ and the green die is a $3$ $3$ or a $4$ $4$ .

If $G = 3 \Rightarrow R = 2$ $G=3 => R=2$
If $G = 4 \Rightarrow R = 1$ $G=4 => R=1$

And so $n (A \cap B) = 2$ $n(A nn B)=2$ , and therefore we have:

$P (A \cap B) = \frac{2}{36} = \frac{1}{18}$ $P(A nn B) = 2/36 = 1/18$

Now, let use calculate $P (B)$ $P(B)$ in a similar fashion:

$P (B) = \frac{n (B)}{n (T)}$ $P(B) = (n(B))/(n(T))$

Where $n (B)$ $n(B)$ is the number of outcomes for which the green die is a $3$ $3$ or a $4$ $4$ . For this outcome we have:

If $G = 3 \Rightarrow R = 1, 2, 3, 4, 5, 6$ $G=3 => R=1,2,3,4,5,6$
If $G = 4 \Rightarrow R = 1, 2, 3, 4, 5, 6$ $G=4 => R=1,2,3,4,5,6$

And so $n (B) = 12$ $n(B)=12$ , and s before $n (T) = 36$ $n(T)=36$ , and so:

$P (B) = \frac{12}{36} = \frac{1}{3}$ $P(B) = 12/36 = 1/3$

And so we can now calculate:

$P (A ∣ B) = \frac{\frac{1}{18}}{\frac{1}{3}} = \frac{3}{18} = \frac{1}{6}$ $P( A | B) = (1/18) / (1/3) = 3/18 = 1/6$

Answer 2 · 2017-04-01T00:34:11

See other answer for a "proper" discussion of how to evaluate conditional probabilities.
The explanation (below) is simply offered as an alternate, quick-and-dirty way of seeing this result.

Explanation:

If green is $3$ $color(green)3$ or $4$ $color(green)4$
and red is $1$ $color(red)1$ , $2$ $color(red)2$ , $3$ $color(red)3$ , $4$ $color(red)4$ , $5$ $color(red)5$ , or $6$ $color(red)6$

${: (,,,color(green)("green"),), (,ul("Sum of"),ul(" | "),ul(color(green)3),ul(color(green)4)), (color(red)("red"),color(red)1," | ",4,color(magenta)5), (,color(red)2," | ", color(magenta)5,6), (,color(red)3," | ", 6,7), (,color(red)4," | ", 7,8), (,color(red)5," | ", 8,9), (,color(red)6," | ", 9,10) :}$

As can be seen from the table:
$color(white)("XXX")$ there are a total of $12$ possible outcomes
and
$color(white)("XXX")$ only $color(magenta)2$ of those outcomes meet the requirement that the total be $5$

So the probability is $2/12 = 1/6$

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Two fair dice (one red and one green) are rolled. What is the probability that the sum is 5, given that the green one is either 4 or 3?

2 Answers

Explanation:

Explanation:

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