Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if:a) a 0 bit and a 1 bit are equally likelyb) the probability that a bit is a 1 is 0.6c) the probability that the i-th bit is a 1 is 1 / (2^i) for i=1,2,3, ...,10

A) 0.0009765625

B) 0.0060466176

C) 2.7756 x 10^ (-17)

Step-by-step explanation:

A) This problem follows a binomial distribution. The number of successes among a fixed number of trials is; n = 10

If a 0 bit and 1 bit are equally likely, then the probability to select in 1 bit is; p = 1/2 = 0.5

Now the definition of binomial probability is given by;

P (K = x) = C (n, k) •p^ (k) • (1 - p) ^ (n - k)

Now, we want the definition of this probability at k = 10.

Thus;

P (x = 10) = C (10,10) •0.5^ (10) • (1 - 0.5) ^ (10 - 10)

P (x = 10) = 0.0009765625

B) here we are given that p = 0.6 while n remains 10 and k = 10

Thus;

P (x = 10) = C (10,10) •0.6^ (10) • (1 - 0.6) ^ (10 - 10)

P (x=10) = 0.0060466176

C) we are given that;

P ((x_i) = 1) = 1 / (2^ (i))

Where i = 1,2,3 ..., n

Now, the probability for the different bits is independent, so we can use multiplication rule for independent events which gives;

P (x = 10) = P ((x_1) = 1) •P ((x_2) = 1) •P ((x_3) = 1) ••P ((x_4) = 1) •P ((x_5) = 1) •P ((x_6) = 1) •P ((x_7) = 1) •P ((x_8) = 1) •P ((x_9) = 1) •P ((x_10) = 1)

This gives;

P (x = 10) = [1 / (2^ (1)) ]•[1 / (2^ (2)) ]•[1 / (2^ (3)) ]•[1 / (2^ (4)) ] ... •[1 / (2^ (10)) ]

This gives;

P (x = 10) = [1 / (2^ (55)) ]

P (x = 10) = 2.7756 x 10^ (-17)