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Intern  Joined: 14 Nov 2011
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Given a spinner with four sections of equal size labeled  [#permalink]

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Question Stats: 70% (01:00) correct 30% (01:06) wrong based on 212 sessions

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 60627
Re: Given a spinner with four sections of equal size labeled  [#permalink]

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3
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.
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Re: Given a spinner with four sections of equal size labeled  [#permalink]

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1
1
nks2611 wrote:
akrish1982 wrote:
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.

hii ,
karishma , can you guide me where am i wrong in these solutions ...
1 p(A or B) = 1/4+1/4-1/4*1/4= 7/16 now we can calculate not getting A that will be 1-7/16=9/16 as it is the answer
2. to me why is this wrong this method 1-1/4*1/4=15/16 for the above problem just because we are asked to calculate the probability of not getting A on both the tires,that's why and this solution will work if we are asked to calculate at least one . let me know if i am missing anything.../

thanks
In advance If you are talking about the result after spinning two times, the probability will be multiplied. You spin once, AND you spin again. So it is not P(A or B). It is P(A and B)

1/4 is the probability of getting A on a spin.

(1/4)*(1/4) is the probability that you will get A on BOTH spins.

Hence, 1 - (1/4)(1/4) = 15/16 is the probability the you will not get A on both spins. You could get A on one spin though.
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Re: Given a spinner with four sections of equal size labeled  [#permalink]

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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

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).

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

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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"?
Intern  Joined: 03 Apr 2012
Posts: 22
Re: Given a spinner with four sections of equal size labeled  [#permalink]

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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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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 ?
Intern  Joined: 07 Feb 2008
Posts: 33
Location: United States
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Re: Given a spinner with four sections of equal size labeled  [#permalink]

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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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Posts: 10010
Location: Pune, India
Re: Given a spinner with four sections of equal size labeled  [#permalink]

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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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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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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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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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Joined: 26 Aug 2016
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Re: Given a spinner with four sections of equal size labeled  [#permalink]

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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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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]

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akrish1982 wrote:
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.

hii ,
karishma , can you guide me where am i wrong in these solutions ...
1 p(A or B) = 1/4+1/4-1/4*1/4= 7/16 now we can calculate not getting A that will be 1-7/16=9/16 as it is the answer
2. to me why is this wrong this method 1-1/4*1/4=15/16 for the above problem just because we are asked to calculate the probability of not getting A on both the tires,that's why and this solution will work if we are asked to calculate at least one . let me know if i am missing anything.../

thanks
In advance GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4226
Re: Given a spinner with four sections of equal size labeled  [#permalink]

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Top Contributor
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

Since the 4 sections each have EQUAL sizes, P(getting A on a single spin) = 1/4
So, P(NOT getting A on a single spin) = 1 - 1/4 = 3/4

What is the probability of NOT getting an A after spinning the spinner two times?
P(neither spin lands on A) = P(no A on 1st spin AND no A on 2nd spin)
= P(no A on 1st spin) x P(no A on 2nd spin)
= 3/4 x 3/4
= 9/16
= B

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

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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

The probability of not getting an A each spin is ¾. Therefore, the probability of not getting an A twice in two spins is ¾ x ¾ = 9/16.

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