I have a question about probability. I need to know how to solve it not just the answer.?

Question

I need to learn how to solve the question. If it ends up with the answer fine, but I need to know how to solve this kind of question because I have a couple more questions like it and I actually want to learn.

The question:
Pretend you are drawing cards without replacement from the infamous "Iraq's Most Wanted" deck issued by the U.S. Military. If you are drawing from the full deck of 55 cards, what is the probability that you draw three cards, which end up being Saddam Hussein and his two sons.

(The order doesn't matter, its 3 out of 55). My question is for a question like this do I take 55 + 54 + 53 = 1:162 since with each card drawn without replacement the number of cards in the pool changes from 55 to 54 to 53 or how would I calculate it?

If the question were "you draw 30 cards and none of them were saddam hussein" how would something like that be solved? Same way? Explain.

Thanks!!

P.S. I'll give you full credit when I submit my assignment, as a "tutor" of sorts.

Answer 1

yap,, so far kitiany is the only one who's given you the correct answer.

i don't wanna be redundant or anything, but i'm sorta displeased with these answers. i'm a big fan with being able to make a logical connection between how probability works and why it works. a lot of people just shoot out formulas, and that's fine if your only goal is to pass exams, but it just doesn't cut if you want a clear understanding of WHY probability works the way that it does. so i'm just gonna spell this out in english words,,,, just in case all the numbers and symbols don't satisfy your desire for an actual understanding.

if you have 55 cards, then you to draw any given card you'd expect a 1 in 55 chance, right? well, since there are actually 3 cards that will meet the needs, rather than just 1, then this actually makes the job easier. so, the odds that you will draw draw papa hussein OR big brother hussein OR the little hussein on the first draw are actually 3 in 55.

now,,, provided you succeed in your first draw, you will only have 54 cards left and now only 2 cards you're looking for, right? so the odds of drawing either of the remaining 2 husseins out of the remaining 54 cards are 2 in 54.

so there's a 3 in 55 chance of your succeeding on JUST the first draw, and a 2 in 54 chance of your succeeding on JUST the second draw. so what are the odds that you succeed on BOTH draws. well, you have to multiply the fractions. 3/55 x 2/54. i'm not sure if you know how to multiply fractions or if it makes sense to you WHY fractions are multiplied this way, but i'm just gonna give you the formula for this part. all the enumerators (top parts) get multiplied to one another, and all of the denominators get multiplied to each other. 3 x 2 is 6. and 55 x 54 is 2970. so,,,, the odds of succeeding on BOTH the first draw and the second draw are 6 in 2970. keep in mind,, in order to succeed at your final goal, you HAVE TO get both of these draws right. also,,,, you have to get the third one right.

obviously now you'll only have 1 card left that you need so you have to get that exact card. and you'll have 53 remaining cards in the deck (because you've removed 2). so your odds here are 1 in 53. ----- your odds of getting BOTH of the first draws are 6 in 2970, and the odds of getting this third one are 1 in 53. so multiply the fractions. 6 x 1 is still 6, and 2970 x 53 is 157,410. so the odds of getting ALL THREE cards in just 3 draws are 6 in 157,410. or you can say 1 in 26,235. (it's the same thing because we divided both the enumerator and the denominator by 6).

keep in mind,,, these are only the odds at the very beginning before you have drawn any cards. if for example you succeed in your first draw, and THEN ask what the odds are NOW, you will have a new problem with new odds. you will be starting with 54 cards and 2 that you need, and then you will need to get 1 card out of 53. so the odds at THIS POINT are 2/54 x 1/53. and this equals 2/2862. if you simplify it, you get 1 in 1431. the odds of your very first draw (3 in 55) just wouldn't matter since it would be in the past. when dealing with probability, the past standardly irrelevant. (you knew that, right? i just wanted to point out that despite my saying "now once you've succeeded in the first 2 draws.... ", the problem of drawing the three cards only matters BEFORE you've drawn any of them.)

ok,,, now on to your second problem. you wanted to know what the odds are of drawing 30 cards from this deck of 55 and NEVER ONCE drawing saddam hussein, right? i'm gonna show you the LOGICAL way to do this problem, and from it you will learn a very fast trick. it's important though that you understand WHY this fast trick works.

let's start with your first draw. what are the odds that you draw 1 card and it's not hussein? 54 in 55, right? because there are 54 cards that aren't that one. so let's say you succeed in drawing him, what are the odds that you can draw another card that isn't him? well,, there are now 54 cards in the deck, and 53 that aren't him, right? so the odds are 53 in 54. and the odds for drawing a THIRD card that isn't him are 52 in 53.

so,, if your question were very simply, what are the odds of drawing 3 cards that aren't hussein, then the answer would be: 54/55 x 53/54 x 52/53

which equals: 148,824 in 157,410 (because all the enumerators, [54, 53, and 52] when multiplied together equal 148,824. and all of the denominators [55, 54, and 53] equal 157,410.)

you'd probably want to simplify this fraction though, and in order to do that, you'll need to divide both the enumerator and the denominator by the same number. so that means you first have to FIND a number which divides evenly into both sides. right off the bat, you should be able to think of two; 53 and 54. right?? so divide both sides by each of these numbers and you get................ 52 in 55. those numbers look familiar, don't they. the denominator is how many cards there are, and the enumerator is the number of cards minus the number of draws that we did.

so,,, if we look at the first THIRTY cards, then the odds for each should look something like this:

draw #1 - 54/55
draw #2 - 53/54
draw #3 - 52/53
draw #4 - 51/52
draw #5 - 50/51
draw #6 - 49/50
draw #7 - 48/49

and so on, down to,

draw #26 - 29/30
draw #27 - 28/29
draw #28 - 27/28
draw #29 - 26/27
draw #30 - 25/26

now,, if you want, then you can multiply all of the enumerators. and then do all of the denominators. but why bother?? if you see a number on both the left side AND the right side, then you might as well just not multiply it to either. so,,, go ahead and leave 54,,, and 53,,,,,, and 52. in fact,, you'll end up leaving every single number out except for 25 on the left side, and 55 on the right side. and so,,, there will be no multiplication that you need to do. your answer at the end will be 25 in 55, which is "the number of cards minus the draws" over "the number of cards". (and you can further simplify this by dividing both sides by 5, which is 5 in 11.)

i hope this was written clearly. i know it's kinda lengthy, but if you really wanna UNDERSTAND math, then the lengthy way is usually the better way.

=========================================

edit, actually, a couple of other people got the answer in before i did. so kitiany is no longer the only one who got it right.

kircheway also got it right, but for the second part, he gave an explanation for how to 30 cards without drawing ANY OF THE THREE HUSSEINS. if this is what you meant, then listen to his explanation and not mine. it sounded though like what you wanted was just to avoid ONE SINGLE card., in which case, what i wrote is better suited for the problem.

Answer 2

In simple english:
Probability of getting one particular sequence of the 3 cards is 1/(55*54*53). Since any sequence is OK, multiply that probability by the number of ways you can draw the 3 cards, which is 3! (3 factorial). So the probability you get all 3 in any order is 3!/(55 * 54 * 53). This is an example of a Combination formula as distinct from a Permutation formula, which would apply to one particular order and for your example is just 1/55 * 1/54 * 1/53.
On your 2nd question, probability of not getting any of the cards is 52/55 on the 1st draw, 51/54 on the 2nd, etc up to 30 draws. Since every one of these events must happen you multiply the odds of each, getting
52*51*...*23/(55*54*...*26).
Notice all the cancellation you can do, which leaves
25*24*23/(55*54*53). Do you see a general formula emerging here, involving the number of total objects, the number of desired (or undesired) objects, and the number of objects you draw? And a similar formula for the first question?
If you google "permutation combination" you can see how the formulas are commonly written. They use a fraction consisting of a lot of factorials (and resulting cancellations) to express the answer.

Answer 3

For the first question, there are 55*54*53 ways to draw three card, and there are 3*2*1 ways to find SH and sons (because the order does not matter). The probability is therefore

(3*2*1)/(55*54*53) = 1/26235 = 0.0038%

If you want the probability of 30 cards and none of them HS, try work with the opposite problem: what is the chance that HS turns up in 30 card? That is simple: 30 out of 55, or 54.55%. The probability that he does not show up is 100 - 54.55 = 45.45%.

Answer 4

Let us start with the basics.

The first time you draw the probability of pulling out Saddam Hussein's card is 1/55, however without the replacement the next time you try it will be 1/54. The probability will not increase each time you draw it will actually decrease.

If you draw 30 cards then the probability for you to pull the card with Saddam Hussein's is 1/(55-30) = 1/25

Makes sense does it not since with each drawing you are improving your chances?

It incorrect to assume that with each draw the probability of drawing Saddam Hussein's decreases to the contrary it increases.

If we have say 3 decks of say 55, 20, 40 cards each then we have 3 independent events and our probability will be a product of their individual probability

P=(1/55)(1/20)(1/40)=(1/ 44000)

I hope that answers your question.

Answer 5

Since order does not matter as long as you end up with Saddam and his two sons:

Probability on first draw: 3/55 (3 cards out of 55)
Probability on second draw: 2/54
Probability on thrid draw: 1/53
Multiply the above: 6:157,410 or 1: 26,235

Answer 6

Since they had answered the first part, i shall say no more.
As for the 2nd part on the 30 cards chosen but none is Saddam Hussein (SH), it should go like this:
54/55*53/54*52/53*....25/26 = Answer
The logic is that you have the chances of picking 54 cards out of 55 that is not SH. Since you do not put back the 1st card that is picked out, you have 54 cards left and you have the chance of picking 53 cards as ur 2nd card that is not SH. It goes on till the 30th card. Where you will have the probability of picking 25 cards that is not SH out of 26 cards for ur 30th card

Answer 7

The way I see it...

First draw.. you have 3/55 chance of drawing one of the 3 cards, and, if you don't draw one of them.. then it is all over.. so assume you did...

Second draw you have 2/54 chance of drawing one of the remaining 2 cards... again.. if you did not.. it is all over.

Third draw.. you have 1/53 chance of drawing the last card you need... in all.. your chances are (3/55)*(2/54)(1/53) = 6/ 157410 = 1/26235 = 0.000038117

Answer 8

Multiply, not add. 1 : 55 x 54 x 53.

Answer 9

i don't know exactly abt the cards but if all the cards are different than the ans is 1/55*1/54*1/53.
because all the events are independent and for independent events the probability formula goes like :
p1 = probability of event 1
p2 = probability of event 2
p3 = probability of event 3
...

...
pn = probability of event n
then probability of the combined events is p1*p2*p3 ...pn.
hope this helps

Answer 10

Use of mixed strategies introducing probabilistic aspects , so that the payoff to a player has a probability distribution. Determination of optimum strategies is simplified when only some reasonable representative value is considered for a distribution .The distribution mean is used for this purpose in expected-value game theory .Minimize Z( x,x) =h( x) +Q ( z)