Question

If two dice are rolled, what is the probability that the sum of the upturned faces equals$i?$ Find it for$i=2,3,...,11,12.$

Short answer

$$\begin{gathered} Part\left(a\right)\frac{1}{3}\left[2\mathrm{ex}\right] \\ Part\left(b\right)\frac{1}{2} \end{gathered}$$

Step-by-step solution

Given Information.

Consider the given information:

If two dice are rolled and the the value of iis 2, 3, ..., 11, 12.

Explanation.

Considered evens:

$G_1$-the first of the two balls are painted gold.

$G_2$-the second of the two balls is painted gold.

$$\begin{gathered} P\left(G_1\right)=\frac{1}{2}\Rightarrow P\left(G_1^C\right)=\frac{1}{2} \\ P\left(G_2\right)=\frac{1}{2}\Rightarrow P\left(G_2^C\right)=\frac{1}{2} \end{gathered}$$

The evens $G_1$ $\&$$G_2$are independent.

Part (a) Explanation.

$P\left(G_1{G}_2\mid \left(G_1\cup G_2\right)\right)=\frac{P\left[G_1{G}_2\left(G_1\cup G_2\right)\right]}{P\left(G_1\cup G_2\right)}$

$$\begin{gathered} G_1{G}_2\left(G_1\cup G_2\right)=G_1{G}_2 \\ {G}_1\cup G_2={\left(G_1^C{G}_2^c\right)}^c \end{gathered}$$

localid="1648885718980" $$\begin{gathered} P\left(G_1{G}_2\right|(G_1\cup G_2))=\frac{P\left(G_1{G}_2\right)}{P\left[{\left(G_1^C{G}_2^c\right)}^c\right]} \\ =\frac{P\left(G_1\right)P\left(G_2\right)}{1-P\left(G_1^c{G}_2^c\right)}Independence \\ =\frac{P\left(G_1\right)P\left(G_2\right)}{1-P\left(G_1^c\right)P\left(G_2^c\right)}Independence \end{gathered}$$

$$\begin{gathered} =\frac{\frac{1}{2}·\frac{1}{2}}{1-\frac{1}{2}·\frac{1}{2}} \\ =\frac{1}{3} \end{gathered}$$

Explanation.

One specific ball is golden, what is the probability that the other one is golden?

Either $G_1$occurs, and the wanted probability is, localid="1648885736207" $P\left(G_2\mid G_1\right)$or $G_2$occurs, and the wanted probability is$P\left(G_1\mid G_2\right)$.

From the definition of independence,

$P\left(G_1\mid G_2\right)=P\left(G_1\right)=\frac{1}{2}$

$P\left(G_2\mid G_1\right)=P\left(G_2\right)=\frac{1}{2}$

This is different because in $a)$we know that one of the three events happened, and here there are only two possible outcomes - by the color of the second ball.