Chapter 5.3, Problem 101E

  (a)

To determine

Probability of rolling doubles on a single toss of the dice.

Answer to Problem 101E

Probability of rolling doubles on a single toss of the dice is approx. 0.1667.

Explanation of Solution

Given information:

Pair of fair, six − 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 − sided dice:

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

Then

36 possible outcomes of a pair of six − sided dice:

  (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),

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.

  $P\left(\text{Doubles}\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{6}{36}=\frac{1}{6}\approx 0.1667$

  (b)

To determine

Probability for no doubles on first toss, followed by doubles on second toss.

Answer to Problem 101E

Probability for no doubles on first toss, followed by doubles on second toss is approx. 0.1389.

Explanation of Solution

Given information:

Pair of fair, six − 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 − sided dice:

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

Then

36 possible outcomes of a pair of six − sided dice:

  (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),

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.

  $P\left(\text{Doubles}\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{6}{36}=\frac{1}{6}$

Apply the complement rule:

  $P\left(\text{No}\text{doubles}\right)=1-P\left(\text{Doubles}\right)=1-\frac{1}{6}=\frac{5}{6}$

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

  $\begin{array}{l}P\left(\text{No}\text{doubles}\text{followed}\text{by}\text{doubles}\right)=P\left(\text{Doubles}\right)\times P\left(\text{No}\text{Doubles}\right)\\ =\frac{1}{6}\times \frac{5}{6}\\ =\frac{5}{36}\\ \approx 0.1389\end{array}$

Thus,

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

  (c)

To determine

Probability for no doubles on first two tosses, followed by the double on third toss.

Answer to Problem 101E

Probability that the first two tosses are not doubles and third toss is doubles is approx. 0.1157.

Explanation of Solution

Given information:

Pair of fair, six − 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 − sided dice:

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

Then

36 possible outcomes of a pair of six − sided dice:

  (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),

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.

  $P\left(\text{Doubles}\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{6}{36}=\frac{1}{6}$

Apply the complement rule:

  $P\left(\text{No}\text{doubles}\right)=1-P\left(\text{Doubles}\right)=1-\frac{1}{6}=\frac{5}{6}$

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

  $\begin{array}{l}P\left(\text{Twice}\text{no}\text{doubles}\text{followed}\text{by}\text{doubles}\right)=P\left(\text{No}\text{Doubles}\right)\times P\left(\text{No}\text{Doubles}\right)\times P\left(\text{Doubles}\right)\\ =\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\\ =\frac{25}{216}\\ \approx 0.1157\end{array}$

Thus,

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

  (d)

To determine

Probability for the first doubles occurs on the kth toss.

Answer to Problem 101E

Probability for first doubles on kth toss,

  $P\left(\text{First}\text{doubles}\text{on}k\text{th}\right)={\left(\frac{5}{6}\right)}^{k-1}\times \frac{1}{6}$

Explanation of Solution

Given information:

Pair of fair, six − 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 − sided dice:

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

Then

36 possible outcomes of a pair of six − sided dice:

  (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),

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.

  $P\left(\text{Doubles}\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{6}{36}=\frac{1}{6}$

Apply the complement rule:

  $P\left(\text{No}\text{doubles}\right)=1-P\left(\text{Doubles}\right)=1-\frac{1}{6}=\frac{5}{6}$

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

  $\begin{array}{l}P\left(\text{First}\text{doubles}\text{on}k\text{th}\text{toss}\right)=P\left(k-1\text{times}\text{no}\text{Doubles}\text{follwed}\text{by}\text{doubles}\right)\\ =\underset{k-1\text{repetitions}}{\underbrace{P\left(\text{No}\text{Doubles}\right)\times P\left(\text{No}\text{Doubles}\right)\times \mathrm{...}\times P\left(\text{No}\text{Doubles}\right)}}\times P\left(\text{Doubles}\right)\\ =\underset{k-1\text{repetitions}}{\underbrace{\frac{5}{6}\times \frac{5}{6}\times \mathrm{...}\times \frac{5}{6}}}\times \frac{1}{6}\\ ={\left(\frac{5}{6}\right)}^{k-1}\times \frac{1}{6}\end{array}$