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There are nine pieces of paper marked from 1 to 9. They are [#permalink]

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04 Feb 2003, 01:42

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There are nine pieces of paper marked from 1 to 9. They are mixed, and four pieces are drawn at random one by one without replacement and arranged in the order of their drawing. As a result, one has a four-digit number. What is the probability to have this number even?

I have another approach. Use the basic definition of P = Favorable/Total.

Let's find T. There are nine NUMBERED pieces, and four are taken without repetitioin. Order is important, so it is permutations. Therefore, T=A(4,9)=3024

Let's find F. A number is even if and only if its last digit is even. There are four even pieces: 2, 4, 6, 8. Fix the last digit on 2; The first position can be filled with eight cards, the second - with seven, and the third - with six. So, for this particular case we have 8*7*6=336. Since we have four even cards, the total F=4*336=1344

And... P=1344/3024=4/9

The answer is simple, but is is so only for these particular conditions. :new-alien:

I have another approach. Use the basic definition of P = Favorable/Total.

Let's find T. There are nine NUMBERED pieces, and four are taken without repetitioin. Order is important, so it is permutations. Therefore, T=A(4,9)=3024

Let's find F. A number is even if and only if its last digit is even. There are four even pieces: 2, 4, 6, 8. Fix the last digit on 2; The first position can be filled with eight cards, the second - with seven, and the third - with six. So, for this particular case we have 8*7*6=336. Since we have four even cards, the total F=4*336=1344

And... P=1344/3024=4/9

The answer is simple, but is is so only for these particular conditions. :new-alien:

I like this one better, but still the whole thing is a little beyond what I want to know about probabilities. I doubt something like this would show up on the real thing.... unless you were really pushing it there... :robot:

Common guys, the answer is way clother than you think. Off course, your hichtech-mathematical-approach proves it.

Here is my approach:

I does not matter how many cards you draw. Without repetition means that if a card is drawn - it is drawn. It's unique and will not appear again in the row of drawings.

In short, we have 4 even numbers and 5 odd ones. Since you draw the cards without repetition, it does not matter how many cards you draw - the probability that the last number (or any other even number among those drawn) will be 4/9. Which is 4 - favourable outcomes and 9 total.

This approach is also applied to a 52 card deck. The probability that any drawn card is ace of spade is 1/52 (f/t) and the probability that it is hearts is 13/52 (f/t).

Common guys, the answer is way clother than you think. Off course, your hichtech-mathematical-approach proves it.

Here is my approach:

I does not matter how many cards you draw. Without repetition means that if a card is drawn - it is drawn. It's unique and will not appear again in the row of drawings.

In short, we have 4 even numbers and 5 odd ones. Since you draw the cards without repetition, it does not matter how many cards you draw - the probability that the last number (or any other even number among those drawn) will be 4/9. Which is 4 - favourable outcomes and 9 total.

This approach is also applied to a 52 card deck. The probability that any drawn card is ace of spade is 1/52 (f/t) and the probability that it is hearts is 13/52 (f/t).

What do you think

Lynov has the best approach to this problem by seeing it logically rather than resorting to math.

We basically want to know the probability that the 4th card is even. Of course, you can do fancy conditional probability calculations for every possible conditin of the first three cards, then add them up. And you will be absolutely correct and you will have used up 5 minutes of your time.

However, Lynov is quite logical is saying this:

If you have not seen any of the papers, then the probability of ANY ONE paper being even is the same 4/9, whether it be the first, second, thrid, fourth, or one of those still left in the bag. Hence, the prob of the entire number being even = 4/9 since we only care about the 4th number.

Here is another simple way to approach this. I don't think that there is any doubt amongs our readers that the prob for the 1st number drawn to be even is 4/9. Okay, what if we put the first number drawn in the 4th position, the next number in the 3rd position, etc.? Because of the symmetry of the problem, the problem really does not change at all, but this time, it is clear to everyone that the probability that the number in the 4th position is even must be 4/9.

HTH!
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993