Engineering Probability question regarding bits used to test a micro-circuit

AI Thread Summary
The discussion focuses on calculating probabilities related to a random pattern of eight bits, where each bit can be either 0 or 1 with equal likelihood. For part a, the probability that all eight bits are 1 is calculated as (0.5)^8, resulting in approximately 0.003906. The user seeks assistance with parts b to d, which involve using the Binomial distribution to find the probabilities of exactly three bits being 1, at least six bits being 1, and at least two bits being 1. The Binomial formula is introduced as a method to solve these problems, emphasizing the importance of understanding the distribution for accurate calculations. The discussion highlights the need for clarity in applying the Binomial distribution to solve the remaining parts effectively.
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Homework Statement



In a random pattern of eight bits used to test a micro-circuit, each bit is equally likely to be 0 or 1. Assume the values of the bits are independent.


a. What is the probability that all eight bits are 1?
b. What is the probability that exactly three of the bits are 1?
c. What is the probability that at least six of the bits are 1?
d. What is the probability that least two of the bits are 1?

Homework Equations



I'm not really sure, I'm trying to figure this part out. I've tried several different combinations of "formulas" for parts b-d. For b, I tried finding the probability of 3 bits being 1 and 5 being 0 but my calculations failed to yield a correct result. Any help in solving parts b-d is greatly appreciated. Thanks in advance.



The Attempt at a Solution


I've figured out part a. I used P(Bit)=0.5 because it is equally likely to be 0 or 1. And I used P(all = 1)=P(bit)^8=(0.5)^8=.003906.
 
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Do you know the Binomial distribution? If the probability of success is p, then in n independent trials, if X is the number of successes then$$
P(X = k) = \binom n k p^k(1-p)^{n-k}$$You can use that to answer your questions. For example, to verify your answer for all 8 bits = 1, you would calculate$$
P(X=8) = \binom 8 8 \left(\frac 1 2\right)^8\left(1-\frac 1 2\right)^{8-8}$$which agrees with your result.
 
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