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probability

by anjaligeorge1 » Wed Jan 07, 2009 9:46 am
A gardener is going to plant 2 red rosebushes and 2 white rosebushes. If the gardener is
to select each of the bushes at random, one at a time, and plant them in a row, what is the
probability that the 2 rosebushes in the middle of the row will be the red rosebushes?

A. 1/12
B. 1/6
C. 1/5
D. 1/3
E. 1/2
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by mental » Wed Jan 07, 2009 10:00 am
1/6
what is the OA?

Arrangement: RWWR

1st bush to be red: 2/4 = 1/2
2nd to be white = 2/3
3rd to be white = 1/2

1/2*2/3*1/2 = 1/6
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by logitech » Wed Jan 07, 2009 3:59 pm
mental wrote:1/6
what is the OA?

Arrangement: RWWR

1st bush to be red: 2/4 = 1/2
2nd to be white = 2/3
3rd to be white = 1/2

1/2*2/3*1/2 = 1/6
1/6 sounds good to me and the arrangment needs to WRRW eventough it does not change the result :)
LGTCH
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by nervesofsteel » Wed Jan 07, 2009 11:54 pm
Total number of ways to plant 4 rose bushes = 4!

Number of ways of planting red in middle two rows is = 2*2

so prob is 4/(4*3*2*1) = 1/6
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probability

by welcome » Thu Jan 08, 2009 6:35 am
In Such questions, a good idea to see 2 objects appearing togather as one. I mean arrangemeant wrrw can be seen as wRw. now having 3 roses w, R, w will have 3 possible arrangement out of which in one R will come in middle so the probablity of R coming in between = 1/3

Again R can have a prob of 1/2 it self as 2 red roses can be arrnage in 2 ways.

So the resulting prob of two event will be = 1/3 * 1/2 = 1/6.
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by woo » Sun Jan 11, 2009 5:32 am
Can you also use nCr approach?
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