0.12 Probability  (Page 2/8)

Let $E$ represent the event that the sum of the faces of two dice is 7.

Since the possible cases for the sum to be 7 are: (1, 6), (2,5), (3, 4), (4, 3), (5, 2), and (6, 1).

and the probability of the event $E$ ,

$P\left(E\right)=6/\text{36}$ or $1/6$ .

We assume the marbles are $r_1$ , $r_2$ , $r_3$ , $w_1$ , $w_2$ , $w_3$ , $w_4$ , $b_1$ , $b_2$ , $b_3$ . Let the event $C$ represent that the marble is red or blue.

The sample space $S=\left\{r_1,r_2,r_{\mathrm{3,}}{w}_1,w_2,w_3,w_4,b_1,b_2,b_3\right\}$

And the event $C=\left\{r_1,r_2,r_3,b_1,b_2,b_3\right\}$

Therefore, the probability of $C$ ,

$P\left(C\right)=6/\text{10}$ or $3/5$ .

Since two marbles are drawn, the sample space consists of the following six possibilities.

Let the event F represent that the sum of the numbers is four. Then

Therefore, the probability of $F$ is

$P\left(F\right)=2/6$ or $1/3$ .

The sample space, as in [link] , consists of the following six possibilities.

Let the event $A$ represent that the sum of the numbers is at least four. Then

Therefore, the probability of $F$ is

$P\left(F\right)=4/6$ or $2/3$ .

Clearly the ace of hearts belongs to both sets. That is

.

Therefore, the events $E$ and $F$ are not mutually exclusive.

For clarity, we list the elements of both sets.

Clearly, .

Therefore, the two sets are not mutually exclusive.