Question

You are rolling a pair of balanced dice in a board game. Rolls are independent. You land in a danger zone that requires you to roll doubles (both faces show the same number of spots) before you are allowed to play again.

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

A) 25/36

B) 5/36

C) 1/6

D) 1/36

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

A) 2/36

B) 1/36

C) 5/36

D) 6/36

3. 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.

A) 5/216

B) 25/216

C) 10/216

D) 20/216

4.  Now you see the pattern. What is the probability that the first doubles occurs on the fourth toss? On the fifth toss?

Give the general result, that is, what is the probability that the first doubles occurs on the ?th toss?

A) (5/6)?−1(1/6)

B) (?−1)(5/6)?−1(1/6)

C) (5/6)(1/6)?−1

D) (?−1)(5/6)(1/6)

5.  What is the probability that you get to go again within three turns? (Enter your answer rounded to four decimal places.)

probability:

Answer 1

1) P(rolling doubles on first toss) = 6/36 = 1/6

Option-C) 1/6

2) P(no double on first toss and doubles on second toss) = (1 - 1/6) * 1/6 = 5/36

Option-C) 5/36

3) P(first double occurs on 3rd toss) = (1 - 1/6)² * 1/6 = 25/216

Option-B) 25/216

4) P(first double on 4th toss) = (1 - 1/6)³ * 1/6 = 125/1296

P(first double on kth toss) = (5/6)^?−1 * (1/6)

Option-A) (5/6)^?−1 * (1/6)

5) P(First double within 3 turns) = P(first double occurs on 1st toss) + P(first double occurs on 2nd toss) + P(first double occurs on 3rd toss)

                                                   = 1/6 + 5/36 + 25/216

                                                   = 0.4213