Pretend you are drawing cards without replacement from the infamous "Iraq's Most Wanted" deck issued by the U.S. Military before Saddam Hussein and his gang were killed or captured.
If you are drawing from the full deck of 52 cards (no jokers), what are the following probabilities:
a. You draw a card that is not Saddam Hussein
b. You draw three cards, which end up being Saddam Hussein and his two sons (whose pictures were also in the deck of cards.)
c. You draw 13 cards and not one of them is Saddam Hussein
2006-11-22 06:51:56 · 5 answers · asked by Funtravel 1 in Science & Mathematics ➔ Mathematics
a. 51/52
b. 1/52 * 1/51 * 1/50
c. 51/52 * 50/51 * 49/50 * 48/49 * 47/48 * ...... 39/40
Since we are talking about Iraqs most wanted cards, there is only one card for each peron, so this is a different problem than your classical card drawing probability problems.
a. bascially you have a 51 in 52 chance of not pulling Saddam, it's a single pull, so no replacement to consider. It is very likely that you will not pull Saddam.
b. Drawing three specific cards is highly unlikely, although your chances get slightly better each time. 0.0192, 0.0196, and 0.02 respectively.
c. This one is like the first, but you repeat the pulls until you have 13 cards, and since you aren't replacing, your chances of avoiding pulling Saddam get slightly worse as you pull.
Hope I'm right, and I hope this helps.
2006-11-22 07:18:55 · answer #1 · answered by wvukid21 2 · 0⤊ 0⤋
This is a 52-card deck, yes?
a. How many times does Saddam appear in the deck? Let that be s (could be 1). The probability that a single card drawn is not Saddam is
(52 - s) / 52
b. For this, I will assume that Saddam and each of his two sons appear once each in the deck. We need three cards with probabilities 1/52, 1/51, and 1/50. The probability is
(1/52) * (1/51) * (1/50)
c. Again assuming that Saddam appears only once in the deck, we are looking for the probability that he remains somewhere in the 39 cards not drawn. Thus the probability is 39/52, or 3/4.
2006-11-22 06:55:46 · answer #2 · answered by ? 6 · 0⤊ 1⤋
First pay attention of the form between odds and hazard. the opportunities being a million:5 ability the hazard is a million/6 and for 2:13, this is two/15. in view that those 2 activities are mutually unique, you could sum the possibilities which grants 5/30 + 4 / 30 this is 3/10. So the opportunities are 3:7.
2016-10-17 09:51:24 · answer #3 · answered by wysong 4 · 0⤊ 0⤋
A. is the lowest and C. is the highest
2006-11-22 07:00:20 · answer #4 · answered by ? 6 · 0⤊ 1⤋
b. would be the lowest in probability and I think a. would probably be the highest in probability.
2006-11-22 06:56:19 · answer #5 · answered by case_mccormack 1 · 1⤊ 1⤋