# Thread: Repeated Tossing of Balanced Die

1. ## Repeated Tossing of Balanced Die

Hi mathforumers,

I have a question that I was hoping someone could confirm or reject the answer I got. I will type it in its entirety first, then explain how I got my answer. The question is from a first-year university Probability course.

A girl tosses a balanced die repeatedly. Suppose she stops tossing as soon as she observes at least one odd number and one even number. What is the probability that she tosses the die exactly four times?

I conceived of the answer as follows:

The first toss could result in even or odd, so there are two possible paths to reach the fourth toss.

A) She gets an odd number first. So then throw #2 and #3 must also be odd, for the game to continue. On throw #4, an even number finally emerges.

Therefore the probability of this event (Odd, Odd, Odd, Even) is (1/2)^4 = 1/16.

B) She gets an even number first. So then throw #2 and #3 must also be even for the game to continue. On throw #4, an odd number finally emerges.

Same probability, 1/16.

Sum = 2/16 or 1/8. My final answer was 1/8.

Is my reasoning correct?

Thanks!

2. sounds good to me

3. 1/8 is correct.
Reasoning can be shortened:
toss#1: odd or even
toss#2: must be same: 1/2
toss#3: must be same: 1/2
toss#4: must be different: 1/2

(1/2)^3 = 1/8

4. Thanks everyone for their input!