Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The experiment is rolling a pair of dice, two times. Rolling a $7$, $11$, or doubles is a success, rolling anything else is a failure.

What is the probability of succeeding on at least one of the two rolls.

I calculate the probability of succeeding in a one-roll experiment to be $\frac{14}{36}$:

There are six ways to roll a $7$, six ways to roll doubles, and two ways to roll an $11$, out of $36$ possible outcomes.

For two consecutive rolls, I think that the sample space of outcomes would be $36\cdot 36$.

How would you figure out how many outcomes would succeed in a two-roll experiment? What is the general way to figure out things like this?

share|improve this question
add comment

3 Answers

up vote 5 down vote accepted

In answering this type of question, it is often useful to answer the opposite question!

"At least..." immediately suggests multiple routes, with multiple calculations, while "never" is more easily dealt with

So, there is a probability of 22/36 of not succeeding on the first roll of two dice, and again the same probability of not succeeding on the second roll. The probability of failing is thus (22/36)^2. The probability of succeeding is then (1 - 484/1296) or around 0.6265

share|improve this answer
add comment

I agree that in a single roll we have: $P(\mathrm{success})=\frac{14}{36}$. Now, instead of counting outcomes of two rolls, we can calculate the probability of succeding on at least one of the two rolls (assuming independence of the rolls):

$$ P(\text{at least one success in two rolls})=1-P(\text{no successes in two rolls})\\ =1-P(\text{no success in a single roll})^2=1-\left(\frac{22}{36}\right)^2. $$

share|improve this answer
1  
+1 for arithmetic accuracy :-) –  robjohn Jan 14 '13 at 7:49
add comment

The probability of failure each roll is $1-\mathrm{P}(7)-\mathrm{P}(11)-\mathrm{P}(\text{pair})=1-\frac{6}{36}-\frac{2}{36}-\frac{6}{36}=\frac{11}{18}$. The probability of failing twice is $\left(\frac{11}{18}\right)^2$. Thus the probability of succeeding at least once is $$ 1-\frac{121}{324}=\frac{203}{324} $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.