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Q. 101

Expert-verifiedFound in: Page 354

Book edition
6th

Author(s)
Daren Starnes, Josh Tabor

Pages
837 pages

ISBN
9781319113339

You are tossing a pair of fair, six-sided dice in a board game. Tosses are independent. You land in a danger zone that requires you to roll doubles (both faces showing the same number of spots) before you are allowed to play again.

a. What is the probability of rolling doubles on a single toss of the dice?

b. What is the probability that you do not roll doubles on the first toss, but you do on the second toss?

c. What is the probability that the first two tosses are not doubles and the third toss is doubles? This is the probability that the first doubles occurs on the third toss.

d. Do you see the pattern? What is the probability that the first doubles occurs on the kth toss?

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

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

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

Part d. Probability for first doubles on *${k}^{th}$* toss,

$P(Firstdoubleson{k}^{th})={\left(\frac{5}{6}\right)}^{k-1}\times \frac{1}{6}$

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

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

$(1,1),(1,2),(1,3),(1,4,),(1,5),(1,6)\phantom{\rule{0ex}{0ex}}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\phantom{\rule{0ex}{0ex}}(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\phantom{\rule{0ex}{0ex}}(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\phantom{\rule{0ex}{0ex}}(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\phantom{\rule{0ex}{0ex}}(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

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.

$P\left(Doubles\right)=\frac{Numberoffavorableoutcomes}{Numberofpossibleoutcomes}\phantom{\rule{0ex}{0ex}}=\frac{6}{36}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}\approx 0.1667$

$6$ 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:

$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\phantom{\rule{0ex}{0ex}}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\phantom{\rule{0ex}{0ex}}(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\phantom{\rule{0ex}{0ex}}(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\phantom{\rule{0ex}{0ex}}(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\phantom{\rule{0ex}{0ex}}(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

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 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:

$P(Nodoublesfollowedbydoubles)=P\left(Doubles\right)\times P(NoDoubles)\phantom{\rule{0ex}{0ex}}=\frac{1}{6}\times \frac{5}{6}\phantom{\rule{0ex}{0ex}}=\frac{5}{36}\phantom{\rule{0ex}{0ex}}\approx 0.1389$

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 – sided dice:

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

Then

$36$ possible outcomes of a pair of six – sided dice:

$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\phantom{\rule{0ex}{0ex}}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\phantom{\rule{0ex}{0ex}}(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\phantom{\rule{0ex}{0ex}}(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\phantom{\rule{0ex}{0ex}}(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\phantom{\rule{0ex}{0ex}}(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

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.

$P\left(Doubles\right)=\frac{Numberoffavorableoutcomes}{Numberofpossibleoutcomes}\phantom{\rule{0ex}{0ex}}=\frac{6}{36}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}$

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:

$P(Twicenodoublesfollowedbydoubles)=P(Nodoubles)\times P(NoDoubles)\times P(NoDoubles)\phantom{\rule{0ex}{0ex}}=\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\phantom{\rule{0ex}{0ex}}=\frac{25}{216}\phantom{\rule{0ex}{0ex}}\approx 0.1157$

Thus,

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

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