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The chance of rain on any given day in Tel-Aviv is 50%. What

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The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post 12 Nov 2012, 18:29
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A
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The chance of rain on any given day in Tel-Aviv is 50%. What is the probability that it rained in Tel-Aviv on exactly 4 out of 6 randomly chosen days?

A. 15/64
B. 30/64
C. 1/2
D. 1/4
E. 52/64
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post 12 Nov 2012, 23:25
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anon1 wrote:
The chance of rain on any given day in Tel-Aviv is 50%. What is the probability that it rained in Tel-Aviv on exactly 4 out of 6 randomly chosen days?


15/64
30/64
1/2
1/4
52/64

chances of rain on exactly 4 days and not rain on 2 days = (1/2)^4* (1/2)^2 = 1/64
Chosing 4 days out of 6 = 6!/(4!*2!) = 15

chances of rain on exactly 4 days out of 6 days = 15/64

Ans A it is.
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post 13 Nov 2012, 00:53
Thanks Vips, that made sense to me. The problem I had was not recognizing

"choosing 4 days out of 6"
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post 13 Nov 2012, 04:01
anon1 wrote:
The chance of rain on any given day in Tel-Aviv is 50%. What is the probability that it rained in Tel-Aviv on exactly 4 out of 6 randomly chosen days?

A. 15/64
B. 30/64
C. 1/2
D. 1/4
E. 52/64


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Hope it helps.
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post 13 Nov 2012, 08:27
Yes it does, thank you bunuel
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post 13 Nov 2012, 17:38
Vips0000 wrote:
anon1 wrote:
The chance of rain on any given day in Tel-Aviv is 50%. What is the probability that it rained in Tel-Aviv on exactly 4 out of 6 randomly chosen days?


15/64
30/64
1/2
1/4
52/64

chances of rain on exactly 4 days and not rain on 2 days = (1/2)^4* (1/2)^2 = 1/64
Chosing 4 days out of 6 = 6!/(4!*2!) = 15

chances of rain on exactly 4 days out of 6 days = 15/64

Ans A it is.



Chosing 4 days out of 6 = 6!/(4!*2!) = 15

Is this combinations or permutations?
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post Updated on: 15 Nov 2012, 05:59
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Quote:


Chosing 4 days out of 6 = 6!/(4!*2!) = 15

Is this combinations or permutations?


This is from formula of Combinations 6C2=6!/(4!*2!)

From Permutations: RRRRNN (R is rain, N - no rain). It comes to 6! divided by 4!*2!.
The same.
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Originally posted by fukirua on 14 Nov 2012, 22:57.
Last edited by fukirua on 15 Nov 2012, 05:59, edited 1 time in total.
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post 15 Nov 2012, 02:45
probability of getting k results out of n -nCk/2^n

6C4/2^4
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post 07 Apr 2017, 07:32
Event occurring k times out of total n times =

nCk * (p)^k * (1-p) ^n-k

p = 50% rain = 1/2
1-p = 1/2
nCk = 6C4 = 15
(1/2)^6*15=15/64
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post 25 Dec 2018, 19:26
1
p(x)=P(X=x)=nCx p^x(1-p)^(n-x)

\(p(x)=P(X=x)=nCx · p^x(1-p)^{n-x}\)

In case anybody searches for this formula (having blindly used it with a calculator in the past), Vips0000 has shown how to do it without a calculator during the CAT.
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What  [#permalink]

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New post 25 Dec 2018, 22:09
anon1 wrote:
The chance of rain on any given day in Tel-Aviv is 50%. What is the probability that it rained in Tel-Aviv on exactly 4 out of 6 randomly chosen days?

A. 15/64
B. 30/64
C. 1/2
D. 1/4
E. 52/64


Din any body realised, we can also solve it without probability?

Critical Thinking Method:
Total event occurrence possibility:
(0 rainy day / 6 Days ; 1 rainy day / 6 Days ; 2 rainy day / 6 Days ; 3 rainy day / 6 Days ; 4 rainy day / 6 Days ; 5 rainy day / 6 Days ; 6 rainy day / 6 Days ; )

So as per question the probability for 3 rainy day / 6 Days = 50%
Considering this as an universal occurrence; rest possibilities have a probability of 50% in together. So any option will have possibility of at-least <50% chance.
This helps you to get rid of options: B, E & C.

Now, on an average with 6 event occurrence and 50% chances for all you have even chances of 7% for all the events. Now if we even consider 3 events with 0% probability, we have a max of 21% chance which fits our question statement and options A

PS: Folks let me know if the critical thinking approach was OK. As i know it was fast for sure :P
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Re: The chance of rain on any given day in Tel-Aviv is 50%. What   [#permalink] 25 Dec 2018, 22:09
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