rolling three fair dice / flipping a fair coin

1. You are going to roll 3 dice. The first has 6 sides, numbered 1 through 6 the second has 10 sides, numbered 1 through 10; and the third has 20 sides, numbered 1 through 20. If all dice are fair, what is the probability that, upon rolling all 3 dice, you don't see any even numbers?

2. You have to flip a fair coin till you get 2 heads OR 2 tails in a row. What is the probability that you have to flip the coin 2 times? What is the probability that you have to flip the coin 3 times?

Thank you!

Hello, trackgirl000!

At first reading, these problems looked intricate,
. . but they are actually quite simple.

1. You are going to roll 3 dice.
The first has 6 sides, numbered 1 through 6.
The second has 10 sides, numbered 1 through 10.
The third has 20 sides, numbered 1 through 20.

If all dice are fair, what is the probability that, upon rolling all 3 dice,
you don't see any even numbers?

Click to expand...

"No even numbers" means all three are odd numbers.

On die #1: \(\displaystyle P(odd) \:=\:\frac{3}{6} \:=\:\frac{1}{2}\)

On die #2: \(\displaystyle P(odd) \:=\:\frac{5}{10}\:=\:\frac{1}{2}\)

On die #3: \(\displaystyle P(odd)\:=\:\frac{10}{20}\:=\:\frac{1}{2}\)


Therefore: \(\displaystyle \(\text{odd, odd, odd})\:=\:\frac{1}{2}\,\times\,\frac{1}{2}\,\times\,\frac{1}{2}\:=\:\frac{1}{8}\)

2. You have to flip a fair coin till you get 2 heads OR 2 tails in a row.
(a) What is the probability that you have to flip the coin 2 times?
(b) What is the probability that you have to flip the coin 3 times?

Click to expand...

(a) With two flips, there are four outcomes: \(\displaystyle \:HH,\,HT,\:TH,\:TT\)

We get 2 heads or 2 tails in two of the cases.

Therefore: \(\displaystyle \(\text{two flips}) \:=\:\frac{2}{4} \:=\:\frac{1}{2}\)


(b) With three flips, there are eight outcomes:
. . \(\displaystyle HHH,\:HHT,\:HTH,\:HTT,\:THH,\:THT,\:TTH,\:TTT\)

In two cases, we must flip three times: \(\displaystyle HTT,\;THH\;\) **

Therefore: \(\displaystyle \(\text{three flips}) \:=\:\frac{2}{8}\:=\:\frac{1}{4}\)

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**

Note that \(\displaystyle HHH\) and \(\displaystyle HHT\) are actually impossible.
. . (If we flip \(\displaystyle HH\), we'd stop.)
The same is true for \(\displaystyle TTH\) and \(\displaystyle TTT\).

But this does not change the probabilities.
. . My answer above is correct.

If you need an explanation for this, I can provide it.

thank you! i was on the right track for most of them, i just needed clarification