Given a spinner with four sections of equal size labeled : Problem Solving (PS)

ashish8

Intern

Joined: 28 Sep 2011

Posts: 47

Own Kudos [?]: 85 [2]

Given Kudos: 10

Location: United States

GMAT 1: 520 Q34 V27

GMAT 2: 690 Q47 V38

GPA: 3.01

WE:Information Technology (Commercial Banking)

Send PM

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

01 May 2012, 10:31

2

Kudos

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"?

akrish1982

Intern

Joined: 07 Feb 2008

Posts: 31

Own Kudos [?]: 148 [2]

Given Kudos: 14

Location: United States

WE:Consulting (Computer Software)

Send PM

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

03 May 2012, 19:08

2

Kudos

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.

L

KarishmaB

Tutor

Joined: 16 Oct 2010

Posts: 14823

Own Kudos [?]: 64928 [2]

Given Kudos: 426

Location: Pune, India

Send PM

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

11 Mar 2017, 01:48

1

Kudos

1

Bookmarks

Expert Reply

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.

G

colorblind

Manager

Joined: 30 Dec 2015

Posts: 58

Own Kudos [?]: 118 [1]

Given Kudos: 173

GPA: 3.92

WE:Engineering (Aerospace and Defense)

Send PM

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

24 Oct 2016, 20:22

1

Kudos

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"?

L

ScottTargetTestPrep

Target Test Prep Representative

Joined: 14 Oct 2015

Status:Founder & CEO

Affiliations: Target Test Prep

Posts: 18761

Own Kudos [?]: 22055 [1]

Given Kudos: 283

Location: United States (CA)

Send PM

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

16 Sep 2019, 11:32

1

Kudos

Expert Reply

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

L

Bunuel

Math Expert

Joined: 02 Sep 2009

Posts: 92929

Own Kudos [?]: 619098 [1]

Given Kudos: 81609

Send PM

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

02 Dec 2020, 01:01

1

Kudos

Expert Reply

ayerhs wrote:

Hey Bunuel, what is the difference between "NOT getting an A after spinning the spinner two times" and "NOT getting A on both spins"? I am still confused.

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.

1. NOT getting an A after spinning the spinner two times, means not getting A in either of the two spins = 3/4*3/4=9/16. So, the following 9 cases out of 16:
BB
BC
CB
BD
DB
CC
CD
DC
DD

2. NOT getting A on both spins (so no AA) means anything but AA. So, the following 15 case out of 16:
AB
BA
AC
CA
AD
DA
BB
BC
CB
BD
DB
CC
CD
DC
DD

Hope it helps.

jayaddula

Intern

Joined: 03 Apr 2012

Posts: 18

Own Kudos [?]: 25 [0]

Given Kudos: 10

Send PM

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

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?

ashish8

Intern

Joined: 28 Sep 2011

Posts: 47

Own Kudos [?]: 85 [0]

Given Kudos: 10

Location: United States

GMAT 1: 520 Q34 V27

GMAT 2: 690 Q47 V38

GPA: 3.01

WE:Information Technology (Commercial Banking)

Send PM

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

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 ?

L

KarishmaB

Tutor

Joined: 16 Oct 2010

Posts: 14823

Own Kudos [?]: 64928 [0]

Given Kudos: 426

Location: Pune, India

Send PM

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

04 May 2012, 20:55

Expert Reply

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.

B

PiyushK

Director

Joined: 22 Mar 2013

Status:Everyone is a leader. Just stop listening to others.

Posts: 611

Own Kudos [?]: 4595 [0]

Given Kudos: 235

Location: India

GPA: 3.51

WE:Information Technology (Computer Software)

Send PM

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

17 May 2014, 13:07

Meaning is confusing ... after spinning 2 times .. means third time or more.. or spinning spinner by hand 2 times..

L

KarishmaB

Tutor

Joined: 16 Oct 2010

Posts: 14823

Own Kudos [?]: 64928 [0]

Given Kudos: 426

Location: Pune, India

Send PM

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

18 May 2014, 21:36

Expert Reply

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.

B

iqahmed83

Intern

Joined: 26 Aug 2016

Posts: 10

Own Kudos [?]: 55 [0]

Given Kudos: 29

Send PM

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

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;

G

colorblind

Manager

Joined: 30 Dec 2015

Posts: 58

Own Kudos [?]: 118 [0]

Given Kudos: 173

GPA: 3.92

WE:Engineering (Aerospace and Defense)

Send PM

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

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)}

S

nks2611

Manager

Joined: 24 Oct 2016

Posts: 196

Own Kudos [?]: 63 [0]

Given Kudos: 89

Location: India

Concentration: Finance, International Business

Schools: IIMB

GMAT 1: 550 Q42 V28

GPA: 3.96

WE:Human Resources (Retail Banking)

Send PM

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

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

L

BrentGMATPrepNow

GMAT Club Legend

Joined: 12 Sep 2015

Posts: 6818

Own Kudos [?]: 29936 [0]

Given Kudos: 799

Location: Canada

Send PM

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

Updated on: 19 Sep 2021, 13:15

Expert Reply

Top Contributor

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

Originally posted by BrentGMATPrepNow on 17 Dec 2017, 15:01.
Last edited by BrentGMATPrepNow on 19 Sep 2021, 13:15, edited 1 time in total.

G

Basshead

Director

Joined: 09 Jan 2020

Posts: 966

Own Kudos [?]: 223 [0]

Given Kudos: 434

Location: United States

Send PM

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

12 Oct 2020, 02:57

The probability of not getting A on each spin is 3/4 (as the probability of getting A on a spin is 1/4)

3/4 * 3/4 = 9/16.

B

ayerhs

Intern

Joined: 30 May 2020

Posts: 20

Own Kudos [?]: 5 [0]

Given Kudos: 309

Send PM

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

01 Dec 2020, 14:09

Hey Bunuel, what is the difference between "NOT getting an A after spinning the spinner two times" and "NOT getting A on both spins"? I am still confused.

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.

gmatclubot

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

01 Dec 2020, 14:09

GMAT Problem Solving (PS) Questions

Given a spinner with four sections of equal size labeled