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

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Manager
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In a jar there are 3 red balls and 2 blue balls. What is the [#permalink] New post 10 Apr 2008, 07:14
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 ?
Senior Manager
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Re: PROBABILITY [#permalink] New post 10 Apr 2008, 07:50
use Formula

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

for the rain question
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Re: PROBABILITY [#permalink] New post 10 Apr 2008, 08:36
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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Re: PROBABILITY [#permalink] New post 10 Apr 2008, 08:41
yup GMAT , I got the same answer as you using the formula,I think there is a typo
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Re: PROBABILITY [#permalink] New post 11 Apr 2008, 07:05
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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Re: PROBABILITY [#permalink] New post 11 Apr 2008, 22:41
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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Re: PROBABILITY [#permalink] New post 11 Apr 2008, 22:45
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 :)
Re: PROBABILITY   [#permalink] 11 Apr 2008, 22:45
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