# Probability Distribution

Good evening everyone,

I'm faced with another question which I can't seem to be able to find a solution online. Thought I get some directions and advice here.

I have the following distribution, which represents the case of the number of events that happen in a single day.

• When (x=0), P(X=x) = 0.2
• When (x=1), P(X=x) = 0.4
• When (x=2), P(X=x) = 0.3
• When (x=3), P(X=x) = 0.08
• When (x=4), P(X=x) = 0.02

Suppose I want to find the probability that the sample standard deviation is 0 if I randomly selected 4 days, how should I proceed?

I have worked out the

• mean of x to be 1.32
• variance of x to be 0.8976
• standard deviation to be 0.947

• If the sample standard deviation was 0, what would the sample look like? Commented Feb 18, 2014 at 7:12
• hmm.. that the sample will be the same for four days? Commented Feb 18, 2014 at 7:32
• Can you edit the title to be more informative? Commented Feb 20, 2014 at 14:52

The first thing for you to note is that choosing a sample with zero standard deviation is equivalent to choosing a sample where the same number of events happen in all four days. Thus, you are looking for $$P(\text{Same number of events in all 4 days})=P(\text{0 events in all days or 1 event in all days or...})$$ To avoid writing these long text statements each time we let $i\in \{0,1,2,3,4\}$ denote the outcome that there are $i$ events in all chosen days, then the probability above can be written as $$P(0\cup1\cup2\cup3\cup4).$$ The next step is to note that the outcomes $i:=\{\text{i events in all 4 days}\}$ are disjoint, i.e. at most one of them can happen. We also call them "mutually exclusive" outcomes. If A and B denote two such outcomes, it holds that $P(\text{A or B})=P(A)+P(B)$ Now generalize this rule to your case.

Let's solve the simpler problem suggested by @whuber and see if this gives you some further insight:

Problem: Suppose the number of events that happen in a single day, denoted $X$, is either 1 or 0, with respective probabilities $P(X=0)=1/3$ and $P(X=1)=2/3$. If you pick two days at random, what is the probability that the standard deviation of your sample is 0?

Solution: The standard deviation is zero exactly when the same number of events occur in all sampled days (in this case two days). There are 4 possible outcomes when sampling 2 days, namely the following:

1. 0 events on the first day and 0 events on the second day
2. 0 events on the first day and 1 event on the second day
3. 1 event on the first day and 0 events on the second day
4. 1 event on the first day and 1 event on the second day.

So, we see that $$P(SD=0)=P(\text{0 events on both days or 1 event on both days}).$$ Now what is this probability? Because the observations are independent the probability of the first outcome is $P(X=0)P(X=0)=1/9$. Likewise, the probability of the fourth outcome is $P(X=1)P(X=1)=4/9$.

Now we use the above observation that when two outcomes are mutually exclusive, i.e. both can never happen at the same time, we have $$P(\text{0 events on both days or 1 event on both days}))=P(\text{Outcome 1 or Outcome 4}) = P(\text{Outcome1}) + P(\text{Outcome 4}) = 1/9 + 4/9 = 5/9.$$ With this at hand, can you solve your version of the problem? The only thing left is to translate this reasoning to your scenario where you select four days with five possible outcomes each. In essence: count how many outcomes satisfy your criteria of zero standard deviation, compute each such outcome's probability and sum those probabilities.

• Thanks Karl. Really appreciate your guidance. Will it be right for me to say that in this case, the probability will be ZERO? Many thanks once again Karl! =) Commented Feb 18, 2014 at 7:30
• That's not correct: think about what it means that the probability is zero; it basically means it can't happen, right? But it's certainly possible to pick, for example, 4 days with 0 events. Maybe the notation $\cup$ confuses you? For two events A and B, $P(A\cup B)$ means the probability that A or B or both happen. The probability that 0 and 1 and 2 and 3 and 4 happen is indeed zero, but any one of them can happen.
– KOE
Commented Feb 18, 2014 at 7:40
• Thanks Karl. In the case that i'd like to find the p (SD = 0), is there any formulaes that I can use for this? Commented Feb 18, 2014 at 7:56
• Yes, the one I wrote in the answer is exactly that. Let me change the notation in the answer and see if you follow.
– KOE
Commented Feb 18, 2014 at 8:04
• When you're stuck it often helps to simplify the problem. For instance, suppose $\Pr(X=0)=1/3$ and $\Pr(X=1)=2/3$ and you change four days to two days. Because there are only four possible samples it is quick and easy to enumerate them, compute their probabilities, and identify those whose standard deviations are zero. This ought to be enough to show you the point of Karl's approach (+1).
– whuber
Commented Feb 18, 2014 at 8:30