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    pauldaoctorpus's Avatar
    pauldaoctorpus Posts: 12, Reputation: 1
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    #1

    Aug 25, 2010, 10:27 PM
    A fair die is tossed six times.What is the probability of getting exactly two 6's ?
    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.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Aug 26, 2010, 05:50 AM

    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:


    So the final answer is

    P(three 4's in 5 tosses) =

    Now, can you apply this technique to your homework problem?
    pauldaoctorpus's Avatar
    pauldaoctorpus Posts: 12, Reputation: 1
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    #3

    Aug 26, 2010, 06:05 AM
    Quote Originally Posted by ebaines View Post
    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:


    So the final answer is

    P(three 4's in 5 tosses) =

    Now, can you apply this technique to your homework problem?
    Thank you very much !

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