Why Not P(A)*(1-P(A)) for Probability?

[nomadreid]

TL;DR

Probability of picking ace from deck, 4/52, OK. But it's also "probability that the card is an ace and nothing else": i.e. (4/52)*(48/52). Why not?

The summary says it all: why is the probability of an event not calculated by the probability that it is the event AND that it is not any other? Sounds silly, and I am certain the explanation is simple, but I don't see it.

 

[etotheipi]

The event $A$ that the card is an ace, A = $\{A_D, A_H, A_S, A_C \}$, is identical to the event B that the card is not anything other than an ace.$$P(A \cap B) = P(A) = P(B) = \frac{4}{52}$$The 'and' is redundant, you'd just be multiplying by 1.

 

[nomadreid]

Ah. Right. Thanks, etotheipi.

 

[Dale]

Formally $P(A\cap B) = P(A) \ P(B|A)$. In this case we have $P(B|A)=1$ so you can indeed calculate it both ways, you just have to use the right formula.

Also, the probability of nothing else is not 48/52. That is the probability of anything else. The probability of nothing else would be 1-48/52 = 4/52

 

[nomadreid]

Thanks, Dale.
As far as "nothing else", that was poorly phrased; you are right, I meant "anything else."

 

[pasmith]

Summary:: Probability of picking ace from deck, 4/52, OK. But it's also "probability that the card is an ace and nothing else": i.e. (4/52)*(48/52). Why not?

The summary says it all: why is the probability of an event not calculated by the probability that it is the event AND that it is not any other? Sounds silly, and I am certain the explanation is simple, but I don't see it.

An event and its complement are not independent, and nor are an event and the complement of its complement (which is the original event itself).