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Given a spinner with four sections of equal size labeled

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Given a spinner with four sections of equal size labeled [#permalink]

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New post 01 May 2012, 02:28
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Given a spinner with four sections of equal size labeled A, B, C, and D, what is the probability of NOT getting an A after spinning the spinner two times?

A. 15/16
B. 9/16
C. 1/2
D. 1/4
E. 1/8
[Reveal] Spoiler: OA
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 01 May 2012, 02:38
pradeepparihar wrote:
Given a spinner with four sections of equal size labeled A, B, C, and D, what is the probability of NOT getting an A after spinning the spinner two times?

A. 15/16
B. 9/16
C. 1/2
D. 1/4
E. 1/8


The probability of NOT getting an A after spinning the spinner two times is 3/4*3/4=9/16 (so getting any of the remaining 3 letters out of 4).

Answer: B.
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 01 May 2012, 09:31
Bunel, if you don't mind I want to expand on this problem for other probability concepts.

A. If I was to do 1 - (1/4*1/4) = 15/16.....This should be the right answer to what type of question?
C. 1/4 + 1/4 = 1/2, I assume this would be the right answer for "Probability of hitting A on any one of the two tries"?
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 02 May 2012, 15:26
ashish8 wrote:
Bunel, if you don't mind I want to expand on this problem for other probability concepts.

A. If I was to do 1 - (1/4*1/4) = 15/16.....This should be the right answer to what type of question?
C. 1/4 + 1/4 = 1/2, I assume this would be the right answer for "Probability of hitting A on any one of the two tries"?


I am going to give it a shot, hopefully someone can confirm/correct me.
For A - that will be the right answer if the question were, what is the probablity if the hits atleast once if spun twice?
p(e) = 1 - (p(no A at all)

For B - that will be right, if the question were, what is the probality if spinner hit A on first spin or second spin?
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 02 May 2012, 15:34
Jay wouldn't the probability of A at least once be, 1 - Probability of NOT A = 1 - (3/4*3/4) = 1 - 9/16 = 7/16 ?
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 03 May 2012, 11:49
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 03 May 2012, 18:08
ashish8 wrote:
Bunel, if you don't mind I want to expand on this problem for other probability concepts.

A. If I was to do 1 - (1/4*1/4) = 15/16.....This should be the right answer to what type of question?
C. 1/4 + 1/4 = 1/2, I assume this would be the right answer for "Probability of hitting A on any one of the two tries"?


it should be (1 - 1/4) * (1 - 1/4),
not 1 - (1/4*1/4)

this is because both need to happen simultaneousluy, so we need to multiply. if both can happen idepedently, we can add.
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 04 May 2012, 19:55
ashish8 wrote:
C. 1/4 + 1/4 = 1/2, I assume this would be the right answer for "Probability of hitting A on any one of the two tries"?


The right answer is (1/4) + (1/4) - (1/4)*(1/4) = 7/16
and the right question is: What is the probability that you get A at least once? or What is the probability of getting A on any one of the two tries or on both?

You subtract the probability of getting A on both spins because you have double counted it. Think SETS. The first (1/4) includes the probability of 'A on both spins'. The second (1/4) also includes the probability of 'A on both spins'. So you need to subtract it once.
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 17 May 2014, 12:07
Meaning is confusing ... after spinning 2 times .. means third time or more.. or spinning spinner by hand 2 times..
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 18 May 2014, 20:36
PiyushK wrote:
Meaning is confusing ... after spinning 2 times .. means third time or more.. or spinning spinner by hand 2 times..


"Not getting A after spinning 2 times" means not getting A in either of the two spins.
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 24 Oct 2016, 19:22
Bunuel wrote:
The probability of:
Getting A at leas once after spinning the spinner two times is 1-3/4*3/4=7/16;
Getting A only once after spinning the spinner two times is 1/4*3/4+3/4*1/4=6/16;
NOT getting A on both spins (so no AA) is 1-1/4*1/4=15/16.

Hope it's clear.


Hey Bunuel,

what is the difference between "NOT getting an A after spinning the spinner two times" and "NOT getting A on both spins"?
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 04 Jan 2017, 06:35
Hey guys! Don't quite understand the logic on Bunuels post:

Getting A only once after spinning the spinner two times is 1/4*3/4+3/4*1/4=6/16;
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Re: Given a spinner with four sections of equal size labeled [#permalink]

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New post 19 Feb 2017, 19:45
iqahmed83 wrote:
Hey guys! Don't quite understand the logic on Bunuels post:

Getting A only once after spinning the spinner two times is 1/4*3/4+3/4*1/4=6/16;




Either get {"A" in the first spin and "non A" in the second spin} OR get {"non A" in the first spin and "A" in the second spin}

Hence:
{(P of A) & (P of non A)} or {(P of non A) & (P of A)}
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Re: Given a spinner with four sections of equal size labeled   [#permalink] 19 Feb 2017, 19:45
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