rolling three fair dice / flipping a fair coin

trackgirl000

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Jan 30, 2007
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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?

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

On die #1: P(odd)=36=12\displaystyle P(odd) \:=\:\frac{3}{6} \:=\:\frac{1}{2}

On die #2: P(odd)=510=12\displaystyle P(odd) \:=\:\frac{5}{10}\:=\:\frac{1}{2}

On die #3: P(odd)=1020=12\displaystyle P(odd)\:=\:\frac{10}{20}\:=\:\frac{1}{2}


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



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?

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

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

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


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

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

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

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**

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

 
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