Roll a pair of dice until it lands doubles.What is the probability you will get the first doubles within three rolls? (Correct to four decimal places.)What is the probability it will take between 5 and 7 rolls?

Let's start by the total number of possible outcomes. This is 6 for one dice, and 6 for the other. So in total that's 6 x 6 = 36 different outcomes of the dice roll.

The number of ways we can roll a double is 6.

So the probability of rolling a double would be the number of ways to roll a double divided by the total possible outcomes.

This is : [tex]\frac{6}{36}[/tex] = 0.1667

The probability of NOT rolling a double: 1 - 0.1667 = 0.8333

Since the probability of rolling a double is equal to 1 - probability of not rolling a double, we can answer the first question in the following way.

The probability of NOT rolling a double for the first THREE turns: 0.8333 x 0.8333 x 0.8333 = 0.5786

Probability of rolling a double in first THREE turns: 1 - 0.5786 = 0.4124

The probability it will take between 5 to 7 rolls would be:

EQUATION 1: (probability of rolling no doubles in first 4 turns) x (probability of rolling a double in the next three turns)

Since we have already found the probability of rolling a double in three turns to be 0.4124, all we need to find is the probability of NOT rolling a double in the first four turns.

This is: 0.8333 * 0.8333 * 0.8333 * 0.8333 = 0.4822

Plugging this into EQUATION 1 we get: 0.4822 * 0.4124 = 0.1988