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

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Manager
Joined: 28 Feb 2003
Posts: 100
Followers: 1

Kudos [?]: 3 [0], given: 0

A student has to select 3 subjects out of 6 subjects A, B, [#permalink]  03 Mar 2003, 22:50
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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Joined: 04 Dec 2002
Posts: 12743
Location: United States (WA)
GMAT 1: 750 Q49 V42
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Followers: 2635

Kudos [?]: 12493 [0], given: 3832

Re: Probability [#permalink]  04 Mar 2003, 03:16
Expert's post
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 (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.

-=-=-
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

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
CTO
Joined: 19 Dec 2002
Posts: 250
Location: Ukraine
Followers: 19

Kudos [?]: 36 [0], given: 9

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.
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [1] , given: 0

1
KUDOS
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
Director
Joined: 01 Feb 2003
Posts: 854
Followers: 1

Kudos [?]: 35 [0], given: 0

...good reasoning!
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