Cables transmit data in packet of $12$ bits. Probability that a bit is corrupted $0.15$. Probability that packet has no more than $2$ corrupted bits?

Question

Course: Big Data

Question: Transatlantic cables transmit data in packets of $12$ bits. The probability that a bit is corrupted is $0.15$ (independent of the corruption of other bits).
a. What is the probability that a packet has no more than $2$ corrupted bits?
b. If $6$ packets are sent, what is the probability that at least one packet will contain $3$ or more corrupted bits?
c. If $X$ is the number of packets containing $3$ or more corrupted bits, what is the probability that $X$ exceeds its mean by more than two standard deviations?

Hint
a. and b. are different distributions. Remember that $X$ is discrete.

My Approach
Here is what I have

$p= 0.15=\frac{3}{20} \\ k=2, \\ n=12$

So using the formula,
$${12 \choose 2}\cdot \bigg(\frac{3}{20}\bigg)^2 \bigg(\frac{17}{20}\bigg)$$

That's it so far.

Answer 1

a) no more than three means it could be either 2,1 or zero so you would need to use the formula with these x's and then add them together let's call the answer Z

b) the probability of one packet will have at least 3 corrupted data is 1-Z and then we consider it as the new success probability and hence we want the probability of having one or more then our P(x=1,2,3,4,5,6) or 1-P(x=0)

and I'm sorry about c I couldn't figure it out

Answer 2

To begin with, this page has information on how to type in MathJax and will help make your post easier to read.

Second, you write (12/2) in your formula. The binomial coefficient $\binom{12}{2}$ is not the same thing as $\frac{12}{2}$. Do not confuse them.

Third, you write (17/20) but you are missing a power on that term. It should be raised to the power of $N-2$ which in this case is $(\frac{17}{20})^{10}$.

Lastly, this number you calculated after corrections, $\binom{12}{2}(\frac{3}{20})^2(\frac{17}{20})^{10}$, represents the probability of getting exactly two corrupted bits. You wanted to calculate however the probability of getting at most two corrupted bits (which is any of exactly zero, exactly one, or exactly two).

In general, the formula for the Binomial Distribution is the following:

Given $N$ independent Bernoulli trials (with only two outcomes: success and failure) each with a probability of success $p$, letting $X$ be the random variable for number of successes you have $$Pr(X=k)=\binom{N}{k}p^k(1-p)^{N-k}$$

If curious to calculate something like $Pr(X\leq k)$ instead, note that $$Pr(X\leq k) = Pr(X=0)+Pr(X=1)+\dots+Pr(X=k-1)+Pr(X=k)$$

The answer to part (a) then being $\sum\limits_{k=0}^2\binom{12}{k}(\frac{3}{20})^k(\frac{17}{20})^{12-k}$

For part (b), try describing the scenario again as a binomial distribution. What will be the trials? How many of them will there be?

The sending of a single packet. This will happen six times.

What counts as a "success" or a "failure" in this case?

Having three or more corrupted bits can be thought of as failure while having two or fewer can be thought of as success. (or vice versa)

What is the probability of success? of failure?

This was calculated in part (a).

The chance of at least one packet having at least three corrupted bits is then:

$Pr(X\geq 1) = Pr(X=1)+Pr(X=2)+\dots+Pr(X=6)$, or easier to calculate, $Pr(X\geq 1)=1-Pr(X<1)=1-Pr(X=0)$

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For part (c), note what the mean and standard deviations are for a binomial distribution. Calculate these yourself if you haven't ever seen it done before, or consult a table/book if you don't know how to.

$\mu = Np$ and $\sigma^2 = Np(1-p)$

How many successes need to occur to have $X$ exceed its mean by two or more standard deviations?

Remembering that $X$ can only take values $0,1,2,3,4,5,6$ in this case, is $6$ two or more standard deviations away? What about $5$? What about $4$? etc... Then find the probabilities of each using the distribution you found earlier.