Answer:
0.1612
Step-by-step explanation:
We are given that there are 24 binary digits .
We are supposed to to find the probability that exactly half of the digits are 0.
Probability of success p (getting half of 24 i.e. 12 digits is 0 ) = 0.5
Since the sum of the probabilities is 1 .
So, probability of failure q = 1-0.5 = 0.5
Formula : [tex]P(x)=^nC_r {\cdot} (p)^r {\cdot} (q)^{n-r}[/tex]
So, n = 24
r = 12
p = 0.5
q=0.5
Substituting the values we get :
[tex]P(x)=^{24}C_12 {\cdot} (0.5)^{12} {\cdot} (0.5)^{24-12}[/tex]
[tex]P(x)=\frac{24!}{12!\times (24-12)!}(0.5)^{12} {\cdot} (0.5)^{12}[/tex]
[tex]P(x)=2704156\times 0.000244140625\times 0.000244140625[/tex]
[tex]P(x)=0.16118[/tex]
Thus the probability of exactly half of the digits are 0 is 0.1612
Hence Option D is correct .
