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Given a spinner with four sections of equal size labeled
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01 May 2012, 03:28

5

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A

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E

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25% (medium)

Question Stats:

70% (01:01) correct 30% (00:59) wrong based on 187 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?

Re: Given a spinner with four sections of equal size labeled
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01 May 2012, 03:38

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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

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).

Re: Given a spinner with four sections of equal size labeled
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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"?

Re: Given a spinner with four sections of equal size labeled
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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?

Re: Given a spinner with four sections of equal size labeled
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03 May 2012, 12:49

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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.

Re: Given a spinner with four sections of equal size labeled
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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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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
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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
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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.

Re: Given a spinner with four sections of equal size labeled
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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.../

Re: Given a spinner with four sections of equal size labeled
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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
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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) = 3/4x3/4 = 9/16 = B

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