Expert: Josh - 7/26/2005

Question
I see now! Thanks, but I thought they had all to be the same probability, how come they can be different?

Thank you.

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Followup To
Question -
Hi, sorry about this but I've just been thinking about this and:

P("KPW")+P("KWP")+P("WPK")+P("WKP")+P("PWK")+P("PKW")=1

So P represents 0.16r but why is it being timsed by the possible outcomes?

Does each one of them represent one combination so really in the brackets it would be 1?

By the way, this is the one about the outcomes of drawing names from a hat, Kate, Pam & Wendy.

Thank you for an help and the previous help you have given me in the past, you have been a great help!

Thanks!
Answer -
Hi Alastair,

Sorry about the notation. Let me clarify this.
P("KPW") represents the probability of drawing the names Kate, Pam and Wendy in this particular order. In short, Kate, Pam and Wendy is abbreviated by the string "KPW".
The prefix P outside the parenthesis denotes the probability; it has nothing to do with the name Pam, per se. Here, P(.) does not involve any multiplication whatsoever. It is just a standard notation for the probability of seeing a certain outcome; where the outcome is specified inside the parenthesis (as a string, or ordered sequence). For example, P("WPK") represents the probability of drawing the name Wendy, followed by Pam and Kate, in this particular order.

What it means when the probabilities add up to 1
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Since there are six terms in the expression P("KPW")+P("KWP")+P("WPK")+P("WKP")+P("PWK")+P("PKW")=1, there are 6 possible outcomes. The fact that these probabilities add up to 1 implies that we are absolutely certain that we will obtain one of the sequence {"KPW" or "KWP" or "WPK" or "WKP" or "PWK" or "PKW"}. In other words, it is not possible to draw three names out of 3 names in any other order. We have considered all possible outcomes.

In general, however, the events need not occur with the same probability. For instance, the probability of getting "KPW", which we write, P("KPW") might be 0.2 (20%).
P("KWP") might be 0.21 (21%), P("PKW") might be 0.19 (19%), P("PWK") might be 0.10 (10%), P("WKP") might be 0.15 (15%), P("WPK") might be 0.15 (15%). I just made up the numbers here in this example, but it is a legitimate scenario, because the probability of all six events add up to 1.

We can only deduce that P("KPW")=P("KWP")=P("PKW")=P("PWK")=P("WKP")=P("WPK")=1/6 IF we are told that all outcomes are equally likely.

I'm glad that you asked a follow-up.

Cheers.

Answer
The probability for each event may or may not be the same. We have to be careful about making assumptions. Just like tossing a dice, if the dice is fair, then, all events are equally probable. There are plenty of random events which do not occur with equal probability. Suppose the weather can only be described as e1="fine", e2="overcast", e3="rain", e4="snow", e5="thunder", e6="others".
Then, as always, we have P(e1)+P(e2)+...+P(e6)=1.
But in summer, e4="snow" is pretty unlikely. So, you would expect P(e4) to be less than 1/6, for instance.

Knowing the weather today, may also help you predict the weather tomorrow. You may come across something called "conditional probability". This asks the question, what is the chance of B happening, given a condition A.
I'm not familiar with the US education system. In Australia, I believe this stuff is covered in year 9 (for high school students around 15 years of age).