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niko · Aug 29, 2005, 12:52 PM

A bowl contains ten chips numbered 1,2,. 10 respectively. Five chips are drawn at random, one at a time, without replacement. What is the probability that 2 even-numbered chips are drawn and they occur on even-numbered draws?

I tried
(OEOEO)
5/10*5/9*4/8*3/6=5/126

and then I tried all combinations
(OEOEO), (EEEEE), (OEEEE), (OEOEE), (OEEEO), (EEEEO), (EEOEE), (EEOEO)
which was =2/9

What am a doing wrong?

The answer is 1/32.

eawoodall · Oct 3, 2005, 05:20 AM

you ask a lot of probability problems, are you having trouble or is askmehelpdesk just doing all your work for you? Honestly.

I think you forgot there are 5 chips drawn at a time, but you only had four fractions multiplied together, so you forgot one fraction included in each probability.
ex: OEOEO 5/10*5/9*4/8*4/7*3/6 did you not forget the second E?
each term must have a fraction, must it not? That should help.

Originally Posted by niko

A bowl contains ten chips numbered 1,2,....10 respectively. Five chips are drawn at random, one at a time, without replacement. What is the probability that 2 even-numbered chips are drawn and they occur on even-numbered draws?

I tried
(OEOEO)
5/10*5/9*4/8*3/6=5/126

and then I tried all combinations
(OEOEO), (EEEEE), (OEEEE), (OEOEE), (OEEEO), (EEEEO), (EEOEE), (EEOEO)
which was =2/9

What am a doing wrong??

The answer is 1/32.

niko · Oct 8, 2005, 11:26 AM

No askmehelp isn't doing all the work for me. I am not even in a course for this. And you didn't help me at all because I did my calculation with five places.

eawoodall · Oct 8, 2005, 11:47 PM

You are saying that

5/10*5/9*4/8*3/6 is five terms, not four?
Looks like four terms to me.
Maybe you are not counting what I am counting.
1st term * 2nd term *3rd term* 4th term * fifth term.
Seems the fifth term is still missing, perhaps?

CroCivic91 · Oct 9, 2005, 07:55 AM

Originally Posted by eawoodall

you are saying that

5/10*5/9*4/8*3/6 is five terms, not four?
looks like four terms to me.
maybe you are not counting what i am counting.
1st term * 2nd term *3rd term* 4th term * fifth term.
seems the fifth term is still missing, perhaps?

Looks like 5 to me.

5/10*5/9*4/8*3/6 = 5/10*5/9*4/8*3/6*1

eawoodall · Oct 9, 2005, 07:53 PM

Oh sorry, they did not spell out the 1/1 so I was not seeing the last term.

jankin · Nov 14, 2005, 08:31 AM

there is no fifth term... u should calculate the value out the ans for the fifth term is not one it is impossible to get 5/126 with the terms of value she gave
I calculated and it required..
5/10*5/9*4/8*4/7*3/6 = 5/126
only then u will obtain 5/126...