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A student has to select 3 subjects out of 6 subjects A, B,

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A student has to select 3 subjects out of 6 subjects A, B, [#permalink]

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New post 03 Mar 2003, 23:50
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A student has to select 3 subjects out of 6 subjects A, B, C, D, E and F. If he
has already chosen E, what is the probability that he will choose B also?
(1) 0.2
(2) 0.25
(3) 0.4
(4) 0.8

What is the radius of the inscribed circle to a triangle whose sides measure
21cm, 72cm and 75cm respectively?
(1) 9 cm
(2) 37.5 cm
(3) 28.5 cm
(4) 14.5 cm


bhavesh

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Re: Probability [#permalink]

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New post 04 Mar 2003, 04:16
brstorewala wrote:
1) A student has to select 3 subjects out of 6 subjects A, B, C, D, E and F. If he
has already chosen E, what is the probability that he will choose B also?
(1) 0.2
(2) 0.25
(3) 0.4
(4) 0.8

2) What is the radius of the inscribed circle to a triangle whose sides measure
21cm, 72cm and 75cm respectively?
(1) 9 cm
(2) 37.5 cm
(3) 28.5 cm
(4) 14.5 cm


bhavesh



I will try to deal with #2 Logically :wink: (if it works)

But can a triangle with a side of 21 have a circle inscribed into it with a radius more than 10.5? (which by itself is pushing the limits)...

So it would be 9, without calculations.


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New post 04 Mar 2003, 23:50
The first one is a clear conditional probability. What will hapen if something has already happened.

P(EB) = P(E)*P(E/B)

1/30 = 1/6*P(E/B)

P(E/B) = 1/5 = 0.2

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New post 18 May 2003, 07:12
stolyar, i disagree with you.

In the formula P(EB) = P(E) * P(B/E)
P(EB) stands for P of event E and event B happening together, when they're dependent.
P(E) stands for P of event E happening.
p(B/E) stands for P of event B happening on condition that event E has happened.

In this problem,
p(E) = 5C2 / 6C3 = 50%
p(B/E) = 4C1 / 5C2 = 40%
p(EB) = 4C1 / 6C3 = 20%

THis way the formula makes sense:
50% * 40% = 20%.

So, i have
p(B/E) = 40%!

I don't think it's good to use conditional formula here. Simply get 4C1/5C2, and you've got the answer! Remember that the guy has ALREADY chosen E, so we can disregard E as an event. It has already happened.

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New post 18 May 2003, 22:11
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let's employ classical approach

Total=5C2=10
Favorable (B+any out of 4)=1*4C1=4

P=4/10=0.4

Oops, you are right :oops:

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New post 18 May 2003, 22:13
...good reasoning! :cool

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  [#permalink] 18 May 2003, 22:13
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