Yes, those are the correct values for one draw.So I'll spell everything out here, and if one of you fine talented folk can tell me if I'm on the right track or not you will make my day!
I have ten cards turned face down. One has a square on it, the other 9 have circles.
I believe I have a 1/10 or 10% chance of drawing the square.
I believe that I have a 9/10 or 90% chance of drawing a circle card.
That is, you draw a card, look at it, put the card back, then shuffle and draw another. This is called "sampling with replacement".If, after each draw, the cards are shuffled, so that it is an entirely random process once again, entirely separate from the initial draw, what are the odds that after two draws (at least) one of them would have been a square?
No, it is not correct. Since 1/10 is the probability of getting the square, 1/10* 1/10 is the probability that both cards have the square, not just "at least one". You also have to consider the probability that the first card has a square (1/10) and the second card does NOT (9/10). What is the probablity of that? And you have to consider the probability that the first cards does NOT have the square but the second card does not. What is the probability of that?1/10 * 1/10 = 1/100? But that's not right. I know the odds are higher than that.
As a check, if there is a 9/10 chance that a card does not have a square, what is the chance that two do not? Subtract that from 1 to get the probability that NOT both cards do not have the square (which is the same as "at least one does have the square).
Yes, that is exactly correct.9/10 * 9/10 = 81/100 of having drawn nothing but circles? That sounds like it could be right, and if so means that the remainder, 19% is the correct odds for (at least) 1 of my 2 cards having been a square?
Those are the probablities you do NOT get the square in that number of draws. Of course the probability of getting the square on the first draw is 1/10, on the first or second draws is 1/10+ (9/10)(1/10)= (1/10)(1+ 9/10), on the first second or third draws (1/10)(1+ 9/10+ 9^2/10^2), on the first, second, third, or fourth draws is (1/10)(1+ 9/10+ (9/10)^2+ (9/10)^3). Can you see what you get on the "n"th draw? Think "geometric series".9/10 * 9/10 * 9/10 = 729/1000 for 27.1% after 3.
9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 = 387420489 / 1000000000 = 0.387420489 or ~61.25% of after 9 draws having (at least) 1 of them be a square.
9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 * 9/10 = 3486784401 / 10000000000 = 0.3486784401 or ~65.1% after 10 draws
The first part of my question is, am I right?
So you want (1/10)(1+ 9/10+ (9/10)^2+ (9/10)^3+ ...)= 9/10. Again, think "geometric series".The second part of my question is:
How many times would I need to draw to have a 90% chance of having drawn at least 1 square card?