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Darina [25.2K]
4 years ago
9

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
Mathematics
2 answers:
OverLord2011 [107]4 years ago
8 0
<span>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 </span>
swat324 years ago
6 0

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 .

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