# 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
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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

I think you are misunderstanding the question. The question says that you are taking a sample of 50 units and want to know what the probability is that 2 or fewer of that 50 are nonconforming, this is the same as asking what the probability is that 48 or more of the 50 are conforming. For the second form of the question would you argue that if all 50 units are conforming that case should not be incuded in the 48 or more? That is the same as p(x=0).

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