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?
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Do you know something about conditional probability? Can you count number of strings starting with $1$? – Paramanand Singh Aug 05 '20 at 15:36
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Please share your thoughts about the problem in the light of my comments. – Paramanand Singh Aug 05 '20 at 15:37
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I know about the conditional probability but not much bcz i just started studying about probabilities of different types and how to use them – Zaima Asif Aug 06 '20 at 00:13
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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 $\frac{9}{2^9}$.
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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 $$\frac{\text{favourable events}}{\text{total events}}= \frac{{9\choose {1}}}{2^{9}}$$
Anindya Prithvi
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Thanks for ur answer. I wanted to ask if it can be done is this way also. Eg 9!/2!7! As there are nine places where it has 2 one so the remaining place 7 zeros . And total outcomes are 2^10= 1024 – Zaima Asif Aug 06 '20 at 00:14
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@ZaimaAsif No, $9\choose2$ would imply that, you already have 1 at the first position (from left) and you're reserving two more positions for 2 more 1s. Total outcomes cannot be 1024 if any of the bit is fixed , in this case the first is fixed.....Had it been, "a random bit is fixed to 1", you would have done $10\choose1$ to fix a random bit, which would still yield the total cases less than 1024 – Anindya Prithvi Aug 06 '20 at 05:08
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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
