# problem with sets and counting again

• Nov 14th 2008, 08:25 PM
Flamed02
problem with sets and counting again
I am having a lot of problems with this problem, I went to a tutor and showed me what to do, but I still get it wrong. Im on the verge of failing, so i want to get the best possible grades on my homework. Please help me with this problem:

If a die is rolled 40 times, there are 640 different sequences possible. What fraction of these sequences have exactly 8 numbers less than or equal to 3? (Round your answer to the nearest 0.0001.)
• Nov 15th 2008, 03:37 AM
Plato
Quote:

Originally Posted by Flamed02
If a die is rolled 40 times, there are 640 different sequences possible. What fraction of these sequences have exactly 8 numbers less than or equal to 3? (Round your answer to the nearest 0.0001.)

First a correction: there are $6^{40}$ different sequences possible.
On any toss of the die there is a probability of 0.5 that the die shows 3 or less.
There are ${40 \choose 8}$, 40 places choosing 8, to have success.
So the answer: ${40 \choose 8}(0.5)^{40}$
• Nov 15th 2008, 01:15 PM
Flamed02
thank you, my answer was 6.994440355 E-5 What does the E-5 mean?