# Probability Change or No?

Update:

Event with P(A)=.1, say drawing an ace from a deck of 10 cards numbered ace-9. *I ONLY NEED 1 ACE TOTAL* If I do the event 4 times at once (4 decks ace-9) is my probability of getting an ace 34.39%? And if I do the event separately (drawing 1 time each from a deck of ace-9 and 3 decks that don't contain an ace) 4 times, are the chances of getting an ace different than the earlier process? What formulas should I be using?

Relevance

Can you please clarify whether you must draw *exactly* one ace or *at least* one ace?

Also, I'm not clear on the events and the drawings. Are you drawing 4 cards (1 from each deck) and then doing that 4 times?

The wording is unclear.

In the first scenario, it sounds like you just want to get *at least* one ace, but it could come from any of the 4 decks. And you are only drawing once from each deck, not 4 times.

Let's first consider the case that you *don't* get any aces.

P(no aces) = 0.9 x 0.9 x 0.9 x 0.9

= 0.6561

So the opposite probability would be that you get at least one ace.

P(at least one ace) = 1 - P(no aces)

= 1 - 0.6561

= 0.3439 (or 34.39%)

I'm not completely clear on the second scenario, but it seems like all that really matters is the first deck. You will never draw an ace from the decks without aces. So you have a 1/10 chance of getting an ace each time you do this. And since you are doing it 4 times, it is exactly equivalent to the first scenario. In order to *not* draw an ace, you'd have to fail to draw an ace from the first deck, and you'd fail 4 times.

P(no ace) = 0.9 x 0.9 x 0.9 x 0.9 = 0.6561

Thus the probability of drawing at least one ace is the opposite probability:

P(at least one ace) = 1 - 0.6561

= 0.3439 (or 34.39%)

It appears that both scenarios have the same probability.

• Brian3 years agoReport

Sorry to ask another , but
If I can only choose 1 of the 4 cards I drew 1st scenario (4 as in 1 from each deck for a total of 4) I only able to pick a max of 1 ace.
2nd scenario, I can get up to 4 aces if I draw an ace each time.
P(being able to get 2)? 3? 4? how likely am i to get more than 1

• Anonymous
3 years ago

No change