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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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 $\displaystyle S$ is $\displaystyle \|S\|=\binom{10}{6}$.Quote:
Originally Posted by Jskid
Of those there $\displaystyle \binom{7}{4}$ 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.
Hello, Jskid!
Quote:
Let $\displaystyle S$ 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: .$\displaystyle \frac{10!}{4!\,6!} \,=\,210$
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.
. . $\displaystyle \_\;1\;\_\;1\;\_\;1\;\_\;1\;\_$
Then I placed 0's in the three interior space:
. . $\displaystyle \_\;1\;0\;1\;0\;1\;0\;1\;\_$
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: .$\displaystyle 5\cdot4 \,=\,20$ ways.
The three can go into separate spaces: .$\displaystyle {5\choose3} \,=\,10$ ways.
Hence, there are:.$\displaystyle 5 + 20 + 10 \,=\,35$ ways with no consecutive 1's.
I'm gratified that I got Plato's answer . . .
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