Tammy is playing a game where she is tryipg to roll a three with a standard die. if she gets a three in any of her first 4 rolls, she wins; otherwise she loses. what is the probability that tammy wins the game? round your answer to the nearest tenth of a percent.

We can partition the events as follows:

Let's compute each term.

Winning at round 1

To win with the first die, Tammy must roll a 3. This happens with probability 1/6. So, we have

Winning at round 2

To win with the second die, Tammy must not roll a 3 with the first roll (probability 5/6), and then roll a 3 with the second roll (probability 1/6). So, we have

Winning at round 3

To win with the third die, Tammy must not roll a 3 with the first two rolls (probability 5/6), and then roll a 3 with the third roll (probability 1/6). So, we have

Winning at round 4

To win with the fourth die, Tammy must not roll a 3 with the first three rolls (probability 5/6), and then roll a 3 with the fourth roll (probability 1/6). So, we have

So, we have

Note

You can do this exercise more quickly by observing

And you lose if you never roll a 3 in 4 rolls (each failure has a probability of 5/6). In fact, you have

step-by-step explanation:

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3