Question

In a randomly generated sequence of 24 binary digits (0s and 1s), what is the probability that exactly half of the digits are 0? a. 1352078 b. 1.295 c. 0 d. 0.1612 Please select the best answer from the choices provided A B C D

Answer 1

P(X=12) = 24C12 * (0.5)^12 * (0.5)^12
= 0.161 to 3 d.p.

not entirely sure, but i think it's that

Answer 2

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 : $P(x)=^nC_r {\cdot} (p)^r {\cdot} (q)^{n-r}$

So, n = 24

r = 12

p = 0.5

q=0.5

Substituting the values we get :

$P(x)=^{24}C_12 {\cdot} (0.5)^{12} {\cdot} (0.5)^{24-12}$

$P(x)=\frac{24!}{12!\times (24-12)!}(0.5)^{12} {\cdot} (0.5)^{12}$

$P(x)=2704156\times 0.000244140625\times 0.000244140625$

$P(x)=0.16118$

Thus the probability of exactly half of the digits are 0 is 0.1612

Hence Option D is correct .