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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, 03:28
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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
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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, 12:49
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Re: Given a spinner with four sections of equal size labeled  [#permalink]

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New post 11 Mar 2017, 01:48
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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New post 01 May 2012, 03:38
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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, 10:31
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, 16:26
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, 16:34
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, 19:08
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, 20:55
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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New post 17 May 2014, 13:07
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, 21:36
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, 20:22
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, 07:35
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, 20:45
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]

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New post 10 Mar 2017, 22:45
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 :)
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Re: Given a spinner with four sections of equal size labeled  [#permalink]

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New post 17 Dec 2017, 15:01
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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


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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New post 16 Sep 2019, 11:32
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 each spin is ¾. Therefore, the probability of not getting an A twice in two spins is ¾ x ¾ = 9/16.

Answer: B
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Re: Given a spinner with four sections of equal size labeled   [#permalink] 16 Sep 2019, 11:32
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