Probability of bits help

cahiersujet

1. Homework Statement
Find the probability that a randomly generated bit string of length 10 begins
with a 1 or ends with 00 if the probability that a bit is a 0 is 0.4.

2. Homework Equations

3. The Attempt at a Solution
The probability that a bit is a 0 is 0.4 and that a bit is a 1 is 0.6.
1 - - - - - - - 0 0
I'm not sure how to move on?

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Wretchosoft

Re: Probability

In this case, it's easier to compute the probability that the bit string will neither begin with 1 nor end with two 0s.

Mark44

Mentor
Re: Probability

Your probability is P(bit string begins with a 1 OR bit string ends with 00). Seems to me that these are independent events, so you can break this probability into the sum of two probabilities, P(bit string begins with a 1) + P(bit string ends with 00).

Hopefully you can just do some fairly simple counting to come up with the two probabilities. For the first one, how many of the 2^10 bit strings start with a 1? For the second one, how many of the 2^10 bit strings end with 00?

gravityandlev

Re: Probability

The other posts give a pretty good start.
Here's a complete solution, for posterity's sake:

The probability that the first digit is a 1 is $latex P_1 = 0.6$.
The probability the the last two digits are 00 is $latex P_{00} = (0.4)^2$.

The probability that the string either begins with a 1 or ends with 00 is given by $latex P_1 + P_{00} - P_1 P_{00} = 0.664$. (That's the probability that you get a 1 in the beginning, plus a probability that you get 00 in the end, minus the probability that both happens. The subtraction at the end avoids double-counting the possibility of both happening).

You'll notice that we never used the number of digits in the string. That's because if we don't care what values they take, then it doesn't matter. The problem would be the same for a 3-digit string.

gravityandlev

Re: Probability

Oops! Here it is, sans formatting errors:

The probability that the first digit is a 1 is $$P_1 = 0.6$$.
The probability the the last two digits are 00 is $$P_{00} = (0.4)^2$$.

The probability that the string either begins with a 1 or ends with 00 is given by $$P_1 + P_{00} - P_1 P_{00} = 0.664$$. (That's the probability that you get a 1 in the beginning, plus a probability that you get 00 in the end, minus the probability that both happens. The subtraction at the end avoids double-counting the possibility of both happening).

You'll notice that we never used the number of digits in the string. That's because if we don't care what values they take, then it doesn't matter. The problem would be the same for a 3-digit string.

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