Answer:
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)
, then it would be: