# The meaning of P(X=0) in a binomial distribution

recently I was resolving an elementary problem about binomial distribution, this problem requests to determinate $P(X<=2)$, where $X$ is nonconforming products, sample is 50 units, and the fraction nonconforming is 0.02. The proposed solution is:
$$P(X\le2) = P(X=0) + P(X=1) + P(X=2).$$ I don’t agree this solution, I think it should be:
$$P(X\le2) = P(0<X<=2) = P(X=1) + P(X=2),$$ because $X=0$ represents “NO nonconforming units”. This is at certain way confirmed when in another problem is requested “determinate probability it will contain at least 1 nonconforming product” and the proposed solution is: $P(X\ge1)=1-P(X=0)$. What are your thoughts about this?

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The question is not clear, why can't you have 0 nonconforming products among the 50 units? –  madness Oct 14 '12 at 22:33
Imagine X to not be the number of nonconforming units, but of heads in coin tosses. You toss 5 times. The probability of heads to come up is 0.5. Now, wouldn't you say it is imaginable that you do not toss a head? Sure it is. So that's your P(x=0). Now swap the 5 for 50, the 0.5 for 0.02 and you see that there indeed is a possibility to get 0 heads and thus you need to consider it when calculating P(x=<2). The same holds for the nonconforming units, which shows the beauty of probability and stats: the essentially same ideas are applicable to so many different examples. –  Momo Oct 14 '12 at 22:40
X=0 certainly represents what you think it does (no nonconforming units) -- but "X=0" is certainly included in "X<=2". Why would you think otherwise? –  Glen_b Oct 14 '12 at 22:48
The part about "at least one" means that the minimum number of units must be 1, so it is all but P(x=0). And since probabilities sum up to 1, P(x>=1) is 1-P(x=0) –  Momo Oct 14 '12 at 22:49
Achilles, you appear to be conflating the ideas of "sample" and "population." The fraction nonconforming in the population could be 0.02, but there is no guarantee that a sample of that population will also contain the same fraction of nonconforming units. In principle, the fraction in the sample (if it is a simple random sample with replacement or if the population size is $2500$ or greater) could be any one of $0/50, 1/50, \ldots, 50/50$. –  whuber Oct 15 '12 at 15:38

In some languages, a double negative is an emphatic form of negative instead of being a positive (as it is in Standard English). Is this why you are having trouble with the meaning of "no nonconforming units"? If so, let's change the question around.

$98\%$ of units are conforming units, $2\%$ are not. In a random sample of $50$ units, what is the probability of at least $48$ conforming units?

to which the answer is that if $Y$ denotes the number of conforming units in the sample, then we are interested in $P\{Y \geq 48\} = P\{Y = 48\} +P\{Y = 49\} + P\{Y = 50\}$. If $X = 50-Y$ denotes the number of nonconforming units, then we have that \begin{align} P\{Y \geq 48\} &= P\{Y = 48\} +P\{Y = 49\} + P\{Y = 50\}\\ &= P\{X = 2\} + P\{X = 1\} + P\{X=0\}\\ \end{align}

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Good point Dilip. This clears my doubt in certain way. That's the beauty of Maths. Thanks. –  Achilles Oct 16 '12 at 16:41

$0 \le 2$ so $P(X\le 2) = P(X=0) + P(X=1) + P(X=2)$ seems reasonable.

If the questions was

What is the probability there were two or fewer nonconforming units?

then I would expect the probability there were zero nonconforming units to be included in that.

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Well, that's my point. Why would you include P(X=0) if it denotes "NO nonconforming products" (that is, "conforming products")? it is not requested for the problem. Really I don't understand the point of including P(X=0). –  Achilles Oct 15 '12 at 0:25
@Achilles: $0$ is a possible outcome and is less than $2$ and so should be included. –  Henry Oct 29 '12 at 0:25