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Q. 101

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The Practice of Statistics for AP Examination
Found in: Page 354
The Practice of Statistics for AP Examination

The Practice of Statistics for AP Examination

Book edition 6th
Author(s) Daren Starnes, Josh Tabor
Pages 837 pages
ISBN 9781319113339

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Short Answer

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 kth toss,

P(First doubles on kth)=(56)k-1×16

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Part a. Step 1. 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.

Part a. Step 2. Calculation

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)(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(Doubles)=Number of favorable outcomesNumber of possible outcomes =636 =16 ≈0.1667

Part b. Step 1. Explanation

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 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(Doubles)=Number of favorable outcomesNumber of possible outcomes=636=16

Apply the complement rule:

P(No Doubles)=1-P(Doubles)=1-16=56

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

P(No doubles followed by doubles)=P(Doubles)×P(No Doubles) =16×56 =536 ≈0.1389

Thus,

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

Part c. Step 1. Explanation

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(Doubles)=Number of favorable outcomesNumber of possible outcomes =636 =16

Apply the complement rule:

P(Not Doubles)=1-P(Doubles)=1-16=56

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

P(Twice no doubles followed by doubles)=P(No doubles)×P(No Doubles)×P(No Doubles) =56×56×56 =25216 ≈0.1157

Thus,

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

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