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 inQuote:
Originally Posted by Jskid
is
.
Of those therewhich 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
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 . . .
.
