91影视

Pair of fair, six 鈥� sided dice is tossed.

Tosses are independent.

Rolling doubles lands in a danger zone before getting another chance to play.

$6$possible outcomes of a fair, six 鈥� sided dice:

$1,2,3,4,5,6$

Then

localid="1665048790953" $36$possible outcomes of a pair of six 鈥� sided dice:

$$\begin{gathered} (1,1),(1,2),(1,3),(1,4,),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{gathered}$$

Where,

First coordinate is the outcome of first dice.

Second coordinate is the outcome of second dice.

Thus,

The number of possible outcomes is $36$.

Also note that

$6$of the $36$outcomes result in doubles:

$(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$

Thus,

The number of favorable outcomes is $6$.

When the number of favorable outcome is divided by the number of possible outcome we get the probability.

$$\begin{gathered} P\left(Doubles\right)=\frac{Numberoffavorableoutcomes}{Numberofpossibleoutcomes} \\ =\frac{6}{36} \\ =\frac{1}{6} \\ 鈮�0.1667 \end{gathered}$$

$6$possible outcomes of a fair, six 鈥� sided dice:

$1,2,3,4,5,6$.

Then

$36$possible outcomes of a pair of six 鈥� sided dice:

$$\begin{gathered} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{gathered}$$

Where,

First coordinate is the outcome of first dice.

Second coordinate is the outcome of second dice.

Thus,

The number of possible outcomes is $36$.

Also note that

$6$of the role="math" localid="1665051979157" $36$outcomes result in doubles:

$(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$

Thus,

The number of favorable outcomes is $6$.

When the number of favorable outcome is divided by the number of possible outcome we get the probability.

$P\left(Doubles\right)=\frac{Numberoffavorableoutcomes}{Numberofpossibleoutcomes}=\frac{6}{36}=\frac{1}{6}$

Apply the complement rule:

$P(NoDoubles)=1-P\left(Doubles\right)=1-\frac{1}{6}=\frac{5}{6}$

Since the tosses are independent of each other, apply the multiplication rule for independent events:

$$\begin{gathered} P(Nodoublesfollowedbydoubles)=P\left(Doubles\right)脳P(NoDoubles) \\ =\frac{1}{6}脳\frac{5}{6} \\ =\frac{5}{36} \\ 鈮�0.1389 \end{gathered}$$

Thus,

Probability for no doubles on first attempt, followed by the doubles on second attempt is approx.$0.1389$.

$6$possible outcomes of a fair, six 鈥� sided dice:

$1,2,3,4,5,6$.

Then

$36$possible outcomes of a pair of six 鈥� sided dice:

$$\begin{gathered} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{gathered}$$

Where,

First coordinate is the outcome of first dice.

Second coordinate is the outcome of second dice.

Thus,

The number of possible outcomes is $36$.

Also note that

$6$of the $36$outcomes result in doubles:

$(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$

Thus,

The number of favorable outcomes is $6$.

When the number of favorable outcome is divided by the number of possible outcome we get the probability.

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

Apply the complement rule:

$P(NotDoubles)=1-P\left(Doubles\right)=1-\frac{1}{6}=\frac{5}{6}$

Since the tosses are independent of each other, apply the multiplication rule for independent events:

$$\begin{gathered} P(Twicenodoublesfollowedbydoubles)=P(Nodoubles)脳P(NoDoubles)脳P(NoDoubles) \\ =\frac{5}{6}脳\frac{5}{6}脳\frac{5}{6} \\ =\frac{25}{216} \\ 鈮�0.1157 \end{gathered}$$

Thus,

Probability for no doubles twice, followed by doubles on third attempt is approx.$0.1157$.