I am some how confused about this problem , but what is did is . The probability of getting exactly two 6's is 2 / 6^6.

I won't simply give you the answer, but I will walk you through the process using a similar example. Then you can aply this technique to solve your homework problem.

Suppose you toss a fair die 5 times- what is the probability of getting exactly three 4's?

The way to think through this problem is like this:

1. First, on any one toss what is the probability of getting a 4? That would be 1/6, since there is one way to get a 4 out of six possibilitis.

2. On any one toss, what's is the probability of tossing a non-4(meaning a 1,2,3,5 or 6)? That would be 5/6.

3. If you were to make 5 tosses, what is the probability of having the first three tosses all be 4's, and the next 2 tosses be non-4's? That would be (1/6)^3 x (5/6)^2

4. Finally, since the three 4's can appear in any order out of the 5 tosses, multiply by the number of ways that three 4's out of 5 tosses can be arranged. That number is C(5,3), or:

\frac {5 \times 4 \times 3} {3!} = 10.


So the final answer is

P(three 4's in 5 tosses) = 10 \times (1/6)^3 \times (5/6)^2 = 0.032

Now, can you apply this technique to your homework problem?

I won't simply give you the answer, but I will walk you through the process using a similar example. Then you can aply this technique to solve your homework problem.

Suppose you toss a fair die 5 times- what is the probability of getting exactly three 4's?

The way to think through this problem is like this:

1. First, on any one toss what is the probability of getting a 4? That would be 1/6, since there is one way to get a 4 out of six possibilitis.

2. On any one toss, what's is the probability of tossing a non-4(meaning a 1,2,3,5 or 6)? That would be 5/6.

3. If you were to make 5 tosses, what is the probability of having the first three tosses all be 4's, and the next 2 tosses be non-4's? That would be (1/6)^3 x (5/6)^2

4. Finally, since the three 4's can appear in any order out of the 5 tosses, multiply by the number of ways that three 4's out of 5 tosses can be arranged. That number is C(5,3), or:

\frac {5 \times 4 \times 3} {3!} = 10.


So the final answer is

P(three 4's in 5 tosses) = 10 \times (1/6)^3 \times (5/6)^2 = 0.032

Now, can you apply this technique to your homework problem?

Thank you very much !