From word possession we take 4 letters without repetition,

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From word possession we take 4 letters without repetition, [#permalink]

04 Dec 2003, 22:03

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From word possession we take 4 letters without repetition, what is the probability to have oops?

How do we solve this problem?

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Re: PS [#permalink]

05 Dec 2003, 00:48

csperber wrote:

From word possession we take 4 letters without

repetition, what is the probability to have oops?

How do we solve this problem?

hmm.without repitition??

from what i understand,we could draw poso ,sopo...etc..and still it would

mean the same, thats oops.

its not like we have to draw it in order...like first o ,second o, third p and

fourth s.. :roll:

Step 1. Total # of ways

Draw 4 letters , order is not important..so we use combinations.

10C4 = 210 ways

Step 2. Favorable ways

We can draw "o's" in 1 ways...p in 1 ways...s's in 4C1 ways

so the number of favoralbe events are 1 * 1C1 * 4C1 = 4 favorable

events which give us the required solution.

Required probability = 4/210 = 4/105 = 2/105

Someone verify please.

thanks
praetorian

EDIT : refer stolyar's post, i made a mistake with the bold part...its been corrected now

Last edited by Praetorian on 05 Dec 2003, 03:40, edited 1 time in total.

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[#permalink]

05 Dec 2003, 03:21

praet, a small correction:

OOPS requires 2 Os (2C2), 1 P (1C1), and 1 S (4C1)

1*1*4/210=4/210=2/105

[#permalink] 05 Dec 2003, 03:21

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From word possession we take 4 letters without repetition,

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From word possession we take 4 letters without repetition,

From word possession we take 4 letters without repetition,

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