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

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01 May 2012, 02:28

3

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A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

64% (01:41) correct
36% (00:33) wrong based on 168 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?

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 [#permalink]

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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"?

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

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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?

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 [#permalink]

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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.

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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01 Dec 2013, 05:28

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

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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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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.

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