# (True/False) Basic Probability Theory

1. Mar 11, 2009

### rela

Dear all,

I have a question.

Suppose we have 3 events X,Y,Z defined as having 200 heads, 400 heads & 600 heads obtained in tossing a fair coin for 800 times.

Then, P(Z)=P(X+Y)=P(600)=P(200+400)=P(X)+P(Y)=P(200)+P(400)

The answer is false but I view it otherwise. My argument is based on the idea of the union of 2 events -> P(X U Y) =P(X) + P(Y). Following this line of reasoning, why is the above statement not considered true?

Please kindly elaborate and direct me to the right understanding level.

Regards
Rela

2. Mar 11, 2009

### tiny-tim

Hi Rela!

P(Z) is the probability of exactly 600 heads …

in full, P(Z) = P({600 heads and 200 tails}) …

so is the statement P({600 heads and 200 tails}) = P({400 heads and 400 tails} U {200 heads and 600 tails}) true or false or meaningless?

3. Mar 12, 2009

### rela

Hi Tim,

Many thanks for your prompt revert.

Hmmm...It looks kinda meaningless to me. But I'm just perturbed by the fact that there exists such a rule in which the probability of the union of 2 statistically events A & B is the the sum of their individual probabilities (i.e P(AUB)=P(A)+P(B).

I just feel that I could apply this rule to the problem statement defined earlier since it makes sense mathematically.

Are you able to elaborate more on the circumstance in which I could apply the above rule correctly then?

Regards
Rela

4. Mar 12, 2009

### tiny-tim

Hi Rela!
Yes, that's right … it's meaningless!
The probability of the union of 2 distinct (non-overlappping) events A & B is the the sum of their individual probabilities.

(and the probability of the intersection of 2 independent events A & B is the the product of their individual probabilities )
yes … you could use the rule if X is exactly 200 heads, Y is exactly 400 heads, and Z is exactly either 200 or 400 heads.

(because X and Y are distinct … ie, they don't overlap … and Z is their union)