Thread: 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!

Originally Posted by 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!

The number of bit-strings in is .

Of those there 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!

Let 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: .

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.
. .

Then I placed 0's in the three interior space:
. .

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: . ways.

The three can go into separate spaces: . ways.


Hence, there are:. ways with no consecutive 1's.

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