Probability of bit strings 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?
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 $\frac{9}{2^9}$.
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