# probability

• Jul 31st 2008, 11:16 AM
kas34
probability
my question is: roll a die 3 times. what is the probability of getting at least two 6's?

also: what is the probability of not getting any even number?
• Jul 31st 2008, 11:47 AM
arbolis
Quote:

roll a die 3 times. what is the probability of getting at least two 6's?
When you pitch one dice, the probability to get a $6$ is $\frac{1}{6}$. If you pitch it twice, it will be $\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}$.
Quote:

what is the probability of not getting any even number?
When you pitch one dice, the probability to not get an even number is $\frac{1}{2}$. So when you pitch it twice, the probability to not get any even number is $\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}$.
To understand why it is so, you have to know that the probability of an event is equal to the number of favorable events divided by the number of possible events. For example when I said that you have a probability of $\frac{1}{2}$ of not getting an even number when you pitch one dice, using the formula we have $\frac{3}{6}=\frac{1}{2}$. 3 over 6 because the favorable outcomes are 1,3 and 5. And the possible outcomes are 1,2,3,4,5 and 6.
• Jul 31st 2008, 11:55 AM
janvdl
Quote:

Originally Posted by kas34
my question is: roll a die 3 times. what is the probability of getting at least two 6's?

also: what is the probability of not getting any even number?

Use the binomial probability for both these:

Probability for at least two 6's when rolled 3 times:

${3 \choose 2} \left( \frac{1}{6} \right) ^{2} \left( \frac{5}{6} \right) ^{1} + {3 \choose 3} \left( \frac{1}{6} \right) ^{3} \left( \frac{5}{6} \right) ^{0}$

Not getting any even number for at least 2 out of 3 throws:

${3 \choose 2} \left( \frac{3}{6} \right) ^{2} \left( \frac{3}{6} \right) ^{1} + {3 \choose 3} \left( \frac{3}{6} \right) ^{3} \left( \frac{3}{6} \right) ^{0}$
• Jul 31st 2008, 12:00 PM
Pn0yS0ld13r
Quote:

Originally Posted by arbolis
When you pitch one dice, the probability to get a $6$ is $\frac{1}{6}$. If you pitch it twice, it will be $\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}$.

Actually, I think the answer is $\frac{7}{8}$ because the question asks what is the probability of getting AT LEAST two 6's.

Consider using binomial distribution, Binomial Distribution -- from Wolfram MathWorld

$\displaystyle\sum_{r=1}^{3}\dbinom{3}{r}\left( \frac{1}{2} \right) ^{r} \left ( \frac{1}{2} \right) ^{3-r} = \frac{7}{8}$
• Jul 31st 2008, 12:01 PM
janvdl
Quote:

Originally Posted by Pn0yS0ld13r
Actually, I think the answer is $\frac{7}{8}$ because the question asks what is the probability of getting AT LEAST two 6's.

Consider using binomial distribution, Binomial Distribution -- from Wolfram MathWorld

$\displaystyle\sum_{r=1}^{3}\dbinom{3}{r}\left( \frac{1}{2} \right) ^{r} \left ( \frac{1}{2} \right) ^{3-r} = \frac{7}{8}$

Missed the "at least". Thanks.
• Jul 31st 2008, 12:07 PM
arbolis
kas34, don't mind my answer, I didn't realize there were 3 pitches.
• Jul 31st 2008, 12:09 PM
janvdl
Quote:

Originally Posted by Pn0yS0ld13r
$\displaystyle\sum_{r=1}^{3}\dbinom{3}{r}\left( \frac{1}{2} \right) ^{r} \left ( \frac{1}{2} \right) ^{3-r} = \frac{7}{8}$

However your solution is also not correct since we want at least 2 successes, not just at least 1.