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# In a jar there are 3 red balls and 2 blue balls. What is the

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Manager
Joined: 25 Mar 2008
Posts: 105

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In a jar there are 3 red balls and 2 blue balls. What is the [#permalink]

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10 Apr 2008, 08:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In a jar there are 3 red balls and 2 blue balls. What is the probability og drawing at least one red ball when drawing two consecutive balls randomly.
A.9/10
B.16/20
C.2/5
D.3/5
E.1/2

23. In Rawanda, the chance for rain on any given time is 50%. What is the probability that it rains on 4 out of 7 consecutive days in Rawanda.

A.4/7
B.3/7
C.45/28
D. 4/28
E. 28/35

PLS TRY TO FIGURE IT OUT ?

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Senior Manager
Joined: 02 Dec 2007
Posts: 450

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10 Apr 2008, 08:50
use Formula

nCr * p^r * (1-p) ^ (n-r)

for the rain question

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CEO
Joined: 29 Mar 2007
Posts: 2554

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10 Apr 2008, 09:36
shobuj wrote:
In a jar there are 3 red balls and 2 blue balls. What is the probability og drawing at least one red ball when drawing two consecutive balls randomly.
A.9/10
B.16/20
C.2/5
D.3/5
E.1/2

23. In Rawanda, the chance for rain on any given time is 50%. What is the probability that it rains on 4 out of 7 consecutive days in Rawanda.

A.4/7
B.3/7
C.45/28
D. 4/28
E. 28/35

PLS TRY TO FIGURE IT OUT ?

1: Lets find the number of ways w/ no red.

2/5*1/4 --> 2/20 --> 1/10 Now we want at least 1 red so its just 1-1/10 --> 9/10

A.

2: First off, to increase ur chances of success, eliminate C, obvs incorrect.

hrmmmm im not doin something correct here.

I get 1/(2)^7 = 1/128

Then we have 7!/4!3! ways to have 4 rainy days. or 35 ways.

Should be 35/128.... arghhh

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Senior Manager
Joined: 02 Dec 2007
Posts: 450

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10 Apr 2008, 09:41
yup GMAT , I got the same answer as you using the formula,I think there is a typo

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Director
Joined: 14 Jan 2007
Posts: 775

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11 Apr 2008, 08:05
Balls problem:
R B
R R
B R
Prob = 9/10

Rain Prob:
Prob= 7C4 (1/2)^4 (1/2)^3 =35/128

Please check the question whether something wrong with the choices.

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Intern
Joined: 08 Apr 2008
Posts: 26

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11 Apr 2008, 23:41
vshaunak@gmail.com wrote:
Balls problem:
R B
R R
B R
Prob = 9/10

Rain Prob:
Prob= 7C4 (1/2)^4 (1/2)^3 =35/128

Please check the question whether something wrong with the choices.

I don't think the answer choices are wrong. It sais 3 CONSECUTIVE DAYS out of 7. 35/128 is the answer for any 3 days out o 7. so it''s not the correct answer for the problem

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Intern
Joined: 08 Apr 2008
Posts: 26

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11 Apr 2008, 23:45
Brindusa wrote:
vshaunak@gmail.com wrote:
Balls problem:
R B
R R
B R
Prob = 9/10

Rain Prob:
Prob= 7C4 (1/2)^4 (1/2)^3 =35/128

Please check the question whether something wrong with the choices.

I don't think the answer choices are wrong. It sais 3 CONSECUTIVE DAYS out of 7. 35/128 is the answer for any 3 days out o 7. so it''s not the correct answer for the problem

Sorry..i didn't read the question properly... you guys are right

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Re: PROBABILITY   [#permalink] 11 Apr 2008, 23:45
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