Probability of bit strings

Question

Suppose you pick a bit string of length $10$. Find the probability that the bit string has exactly two $1$'s, given that the string begins with a $1$. Can someone please explain to me how to do it?

Answer 1

Since you know that the first bit is $1$, you just want to find the probability that exactly one of the remaining $9$ digits is $1$. The total number of possible $9$-bit sequences is $2^9$. Of these only $9$ have exactly a single $1$, one for each possible position of the $1$. So the probability is $\frac{9}{2^9}$.

Answer 2

X~The number of 1s in a bit string of length 10 X~B(10,0.5) We just split it by cases since we know first one is already 1. Y~The number of 1s in a bit string of length 9 Y~B(9,0.5) P(Y=1)=0.01757 P(given first is 1 and we want 2 1s)=(0.01757×0.5)/P(X=2) =0.2

Answer 3

There is a nice and clean way to do it. The number of possible bit strings are $2^{9}$. This is because each and every position in the string can be zero or one except the first which can only be one.

Now, you want one out of the nine places in the string to be 1. In order to do that, just select two places using $9\choose 1$.

Now, the probability is rendered as $$\frac{\text{favourable events}}{\text{total events}}= \frac{{9\choose {1}}}{2^{9}}$$