How many bit strings of 12 contain:?

Question

exactly three 1's?
atleast most three 1's?
atleast three 1's?
an equal number of 0's and 1's?

Answer 1

Totl no. of combinations = 2^12 =4096
N=12
P(getting a 1) =1/2
If X=number of the ones reqd
then the probability is 12CX (1/2)^12
(C=combinations)

For exactly 3,put X=3
probability = 220/4096
no. of strings = 220/4096 * 4096 =220

For atleast 3(thatb is,X greater than or equal to 3)
P=1-P(X<3)
P=(1-(P(X=0)+P(X=1)+P(X=2))
=(1-(1+12+66)/4096)
=4017/4096
=4017 strings
......
.......
equal no of 0s and 1s(X=6)

P=924/4096
=924 strings

At the most 3(X less than or equal to 3)
P=(P(X=0)+P(X=1)+P(X=2)+P(X=3))
=(1+12+66+220)/4096
=299/4096
=299 students

Answer 2

Probability is something of a red herring in the earlier answer. All you need are binomial coefficients, or "choose numbers," which I'll write like C(12,3) for "12 choose 3" which is 12! / (3! 9!).

To count 12 bit strings with exactly three 1s, choose three of the positions to have 1s (the rest will have 0s). That can be done in exactly C(12,3) = 220 ways.

At most three 1s means zero, one, two, or three 1s. So that's C(12,0) + C(12,1) + C(12,2) + C(12,3) = 1 + 12 + 66 + 220 = 299 strings.

At least three 1s means three, four, ..., or twelves 1s. The big sum C(12,3) + C(12,4) + ... + C(12,12) gives the correct answer, but it's easier to do with the complement. At least three 1s is the opposite of at most two 1s. Remember that there are 2^12 = 4096 strings. The number with at least three 1s equals 4096 minus the number with at most two 1s: 4096 - (C(12,0) + C(12,1) + C(12,2)) = 4096 - 79 = 4017 strings.

Finally, an equal number of 0s and 1s just means six 1s, so that's C(12,6) = 924 strings.

Answer 3

An equal number of 0's and 1's. It would be written 1100