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    probability that binary number has no consecutive ones

    Let S be the set of all sequences of 6 zeros and 4 ones. Find the probability that a randomly selected element of S has no two consecutive ones. The sample space is 10!
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    Re: probability that binary number has no consecutive ones

    Quote Originally Posted by Jskid View Post
    Let S be the set of all sequences of 6 zeros and 4 ones. Find the probability that a randomly selected element of S has no two consecutive ones. The sample space is 10!
    The number of bit-strings in S is \|S\|=\binom{10}{6}.

    Of those there \binom{7}{4} which have no consecutive ones.
    The reason for this is that we use the zeros as separators:_0_0_0_0_0_0_. So there are seven places for the 1's.
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    Re: probability that binary number has no consecutive ones

    Hello, Jskid!

    Let S be the set of all sequences of 6 zeros and 4 ones.
    Find the probability that a randomly selected element of S has no two consecutive ones.
    The sample space is 10! . No!

    That would be true if we had 10 different objects to arrange.

    Since we have six identical 0's and four identical 1's,
    . . the sample space is: . \frac{10!}{4!\,6!} \,=\,210

    Plato has the best solution.

    I went about it differently ... and found that it required far more work.


    Instinctively, I set about separating the 1's.

    I arranged the four 1's in a row with a space before, after, and between them.
    . . \_\;1\;\_\;1\;\_\;1\;\_\;1\;\_

    Then I placed 0's in the three interior space:
    . . \_\;1\;0\;1\;0\;1\;0\;1\;\_

    Now I must distribute the three remaining 0's.


    They can all go into any of the 5 spaces: 5 ways.

    Two can go into one space, one into another: . 5\cdot4 \,=\,20 ways.

    The three can go into separate spaces: . {5\choose3} \,=\,10 ways.


    Hence, there are:. 5 + 20 + 10 \,=\,35 ways with no consecutive 1's.


    I'm gratified that I got Plato's answer . . .
    .
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