91影视

Roll two fair dice separately. Each die has six faces.

Sample space:

$\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6),\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}$

Roll two fair dice separately. Each die has six faces.

Total number of possible outcomes$=36$

LetA be the event that either a three or four is rolled first, followed by an even number.

Possible outcomes of the event$=\left\{\begin{array}{l}(3,2),(3,4),(3,6)\\ (4,2),(4,4),(4,6)\end{array}\right\}$

Thus, probability either a three or four is rolled first followed by an even number

$$\begin{gathered} P\left(A\right)=\frac{6}{36} \\ P\left(A\right)=\frac{1}{6} \end{gathered}$$

Roll two fair dice separately. Each die has six faces.

Let B be the event that the sum of the two rolls is at most seven.

Possible outcomes of the events

$=\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5)\\ (3,1),(3,2),(3,3),(3,4)\\ (4,1),(4,2),(4,3)\\ (5,1),(5,2)\\ (6,1)\end{array}\right\}$

Total number of outcomes$=21$

Thus the probability that the sum of the two rolls is at most seven is calculated as

$$\begin{gathered} P\left(B\right)=\frac{21}{36} \\ P\left(B\right)=\frac{7}{12} \end{gathered}$$

Roll two fair dice separately. Each die has six faces.

Let A be the event that either a three or four is rolled first, followed by an even number.

Sample space of the event A

$=\left\{\begin{array}{l}(3,2),(3,4),(3,6)\\ (4,2),(4,4),(4,6)\end{array}\right\}$

Let B be the events that sum of the two rolls is at most seven.

Sample space of the event B

$=\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ (2,1),(2,2),(2,3),(2,4),(2,5)\\ (3,1),(3,2),(3,3),(3,4)\\ (4,1),(4,2),(4,3),\\ (5,1),(5,2)\\ (6,1)\end{array}\right\}$

Thus, possible outcomes for $(A鈭�B)=\left\{\begin{array}{l}(3,2),(3,4),\\ (4,2),\end{array}\right\}$

Total number of outcomes$=3$

$$\begin{gathered} P(A鈭�B)=\frac{P\left(AANDB\right)}{P\left(B\right)} \\ P(A鈭�B)=\frac{{\displaystyle \frac{3}{36}}}{{\displaystyle \frac{21}{36}}} \\ P(A鈭�B)=\frac{3}{21} \\ P(A鈭�B)=\frac{1}{7} \end{gathered}$$

In words, localid="1648370537292" $P(A鈭�B)$means that the face drawn is either a three or four followed by an even number, given that the sum of two rolls is at most seven.

Roll two fair dice separately. Each die has six faces.

Let A be the event that either a three or four is rolled first followed by an even number.

Possible outcomes of the event$=\left\{\begin{array}{l}(3,2),(3,4),(3,6)\\ (4,2),(4,4),(4,6)\end{array}\right\}$

Let B be the event that the sum of the two rolls is at most seven.

Possible outcomes of the events

we observe that there are few common outcomes in events A and B

Thus events A and B are not mutually exclusive.

Roll two fair dice separately. Each die has six faces.

Let A be the event that either a three or four is rolled first, followed by an even number.

Possible outcomes of the event$=\left\{\begin{array}{l}(3,2),(3,4),(3,6),\\ (4,2),(4,4),(4,6)\end{array}\right\}$

Total number of outcomes$=6$

Thus, probability either a three or four is rolled first, followed by an even number is calculated as

$$\begin{gathered} P\left(A\right)=\frac{6}{36} \\ P\left(A\right)=\frac{1}{6} \end{gathered}$$

Let B be the event that the sum of the two rolls is at seven.

Possible outcomes of the events

The possible outcomes$=3$

Thus, $P(A鈭�B)$is calculated as

$$\begin{gathered} P\left(AANDB\right)=P(A鈭�B)P\left(B\right) \\ P(A鈭�B)=\frac{P\left(AANDB\right)}{P\left(B\right)} \\ P(A鈭�B)=\frac{{\displaystyle \frac{3}{36}}}{{\displaystyle \frac{21}{36}}} \\ P(A鈭�B)=\frac{3}{21} \\ P(A鈭�B)=\frac{1}{7} \end{gathered}$$

we observe $P(A鈭�B)鈮�P\left(A\right)$

The first condition is not satisfied. Thus, events A and B are not independent.