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# Renee has a bag of 6 candies, 4 of which are sweet and 2 of

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Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink]

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11 Apr 2011, 02:21
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Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?

A. 3/5
B. 7/18
C. 8/15
D. 4/10
E. 5/12

[Reveal] Spoiler:
Here's what I did: I thought why not use combinatorics lol (too brave right? ). So I thought there are 6!/4!2! = 15 ways to pick 2 candies out of the 6.
I need exactly 1 sour so the chance of getting exactly 2 sour was 4/15. 1-4/15 = 11/15. This was SOOO WRONG !!

2/6 x 4/5 = 8/30 (sour fist and then sweet)
2/6 x 4/5 = 8/30 (sweet fist and then sour)

8/30 + 8/30 = 16/30 = 8/15.

Is it just coincidence that I got 15 out of my nCr calculation? Can someone explain to me if this can also be calculated using combinatorics? I think the reason I used the nCr is because during praciticing on some probab. questions I saw more than a few questions being answered using combinatorics and somehow it seemed easier to me..

Thanks you guys, appreciate it!
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Nov 2013, 01:23, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Rene has a bag of 6 candies, probab& combinatorics possible? [#permalink]

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11 Apr 2011, 02:38
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l0rrie wrote:
Hi everyone,

I'll be needing some help AGAIN.. Yeah I know sorry but this is really my weak point..

Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour.
Jack picks two candies simultaneously and at random. What is the chance
that exactly 1 of the candies he has picked is sour?

Here's what I did: I thought why not use combinatorics lol (too brave right? ). So I thought there are 6!/4!2! = 15 ways to pick 2 candies out of the 6.
I need exactly 1 sour so the chance of getting exactly 2 sour was 4/15. 1-4/15 = 11/15. This was SOOO WRONG !!

2/6 x 4/5 = 8/30 (sour fist and then sweet)
2/6 x 4/5 = 8/30 (sweet fist and then sour)

8/30 + 8/30 = 16/30 = 8/15.

Is it just coincidence that I got 15 out of my nCr calculation? Can someone explain to me if this can also be calculated using combinatorics? I think the reason I used the nCr is because during praciticing on some probab. questions I saw more than a few questions being answered using combinatorics and somehow it seemed easier to me..

Thanks you guys, appreciate it!

You were right in calculating Total Possible ways to pick two candies.

P= Favorable/Total
Total = 6C2 = 6*5/2 = 15

Favorable:
Pick 1 sour candy out of 2 sour candy
AND
Pick 1 sweet candy out of 4 sweet candy

2C1*4C1 = 2*4=8

P=8/15

When you use combination method, it is picking all possible cases and the order doesn't matter. Whereas, upon choosing probability method to solve, order matters.

Thus,
Total Probability:
Probability of choosing sour candy first AND Probability of choosing sweet candy
OR
Probability of choosing sweet candy first AND Probability of choosing sour candy

2/6* 4/5 + 4/6 * 2/5
= 8/30 + 8/30
= 16/30
= 8/15

Why: 2/6* 4/5 + 4/6 * 2/5
P=2/6: total sour candies= 2, total candies=6
1 Sour candy Picked.
AND means "*"
Now, total candies=5(because one picked). Sweet candies=4(First picked was sour so 4 sweet candies still left)
P=4/5
2/6 * 4/5

OR Means "+"
4/6: total sweet candies= 4, total candies=6
1 Sweet candy Picked.
AND means "*"
Now, total candies=5(because one picked). Sour candies=2(First picked was sweet so 2 sour candies still left)
P=2/5
4/6 * 2/5

Does that make little sense?
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Re: Rene has a bag of 6 candies, probab& combinatorics possible? [#permalink]

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11 Apr 2011, 03:24
Thank you so much fluke! I totally get it now!

This forum is such an amazing help!
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Re: Rene has a bag of 6 candies, probab& combinatorics possible? [#permalink]

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11 Apr 2011, 04:28
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(2C1 * 4C1)/(6C2)

2 * 4/6 * 5/2

= 8/15
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Re: Rene has a bag of 6 candies, probab& combinatorics possible? [#permalink]

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24 Apr 2011, 20:00
= (2c1 * 4c1 )/ 6 c2

= 8/15
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Re: Rene has a bag of 6 candies, probab& combinatorics possible? [#permalink]

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07 Nov 2013, 17:05
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink]

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08 Nov 2013, 01:24
Expert's post
1
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l0rrie wrote:
Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?

A. 3/5
B. 7/18
C. 8/15
D. 4/10
E. 5/12

[Reveal] Spoiler:
Here's what I did: I thought why not use combinatorics lol (too brave right? ). So I thought there are 6!/4!2! = 15 ways to pick 2 candies out of the 6.
I need exactly 1 sour so the chance of getting exactly 2 sour was 4/15. 1-4/15 = 11/15. This was SOOO WRONG !!

2/6 x 4/5 = 8/30 (sour fist and then sweet)
2/6 x 4/5 = 8/30 (sweet fist and then sour)

8/30 + 8/30 = 16/30 = 8/15.

Is it just coincidence that I got 15 out of my nCr calculation? Can someone explain to me if this can also be calculated using combinatorics? I think the reason I used the nCr is because during praciticing on some probab. questions I saw more than a few questions being answered using combinatorics and somehow it seemed easier to me..

Thanks you guys, appreciate it!

$$P(Sour=1)=\frac{C^1_4*C^1_2}{C^2_6}=\frac{8}{15}$$

OR:

$$P(Sour=1)=2*\frac{4}{6}*\frac{2}{5}=\frac{8}{15}$$, multiplying by 2 as Sour/Sweet can occur in two ways: Sour/Sweet and Sweet/Sour.

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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink]

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13 Nov 2013, 16:27
thanks for the help !
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink]

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05 Jan 2014, 06:43
Bunuel wrote:
$$P(Sour=1)=\frac{C^1_4*C^1_2}{C^2_6}=\frac{8}{15}$$

OR:

$$P(Sour=1)=2*\frac{4}{6}*\frac{2}{5}=\frac{8}{15}$$, multiplying by 2 as Sour/Sweet can occur in two ways: Sour/Sweet and Sweet/Sour.

Thanks Bunuel for solution.

I got the first method but I have a query regarding second method. Could you please elaborate why 2 should be divided by 5 ? Problem says simultaneously and not one after the other.
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink]

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05 Jan 2014, 06:48
Bunuel wrote:
$$P(Sour=1)=\frac{C^1_4*C^1_2}{C^2_6}=\frac{8}{15}$$

OR:

$$P(Sour=1)=2*\frac{4}{6}*\frac{2}{5}=\frac{8}{15}$$, multiplying by 2 as Sour/Sweet can occur in two ways: Sour/Sweet and Sweet/Sour.

Thanks Bunuel for solution.

I got the first method but I have a query regarding second method. Could you please elaborate why 2 should be divided by 5 ? Problem says simultaneously and not one after the other.

Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink]

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28 Jun 2014, 06:17
Bunuel wrote:
Bunuel wrote:
$$P(Sour=1)=\frac{C^1_4*C^1_2}{C^2_6}=\frac{8}{15}$$

OR:

$$P(Sour=1)=2*\frac{4}{6}*\frac{2}{5}=\frac{8}{15}$$, multiplying by 2 as Sour/Sweet can occur in two ways: Sour/Sweet and Sweet/Sour.

Thanks Bunuel for solution.

I got the first method but I have a query regarding second method. Could you please elaborate why 2 should be divided by 5 ? Problem says simultaneously and not one after the other.

Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.

i solved this by prob of getting atleast i sour = ( 1- prob of getting both sweet) = ( 1- 4c2/6c2) ..............where am i going wrong
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink]

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28 Jun 2014, 07:06
tyagigar wrote:
Bunuel wrote:

Thanks Bunuel for solution.

I got the first method but I have a query regarding second method. Could you please elaborate why 2 should be divided by 5 ? Problem says simultaneously and not one after the other.

Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.

i solved this by prob of getting atleast i sour = ( 1- prob of getting both sweet) = ( 1- 4c2/6c2) ..............where am i going wrong

We are asked to find the probability that exactly 1 of the candies he has picked is sour not at least 1.
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Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are [#permalink]

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16 May 2015, 03:51
Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?

A. 16/225
B. 8/30
C. 7/15
D. 8/15
E. 16/15
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink]

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16 May 2015, 04:49
reto wrote:
Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?

A. 16/225
B. 8/30
C. 7/15
D. 8/15
E. 16/15

Please search before posting. Thank you.
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink]

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28 May 2016, 11:46
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink]

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28 May 2016, 19:07
Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?

A. 3/5
B. 7/18
C. 8/15
D. 4/10
E. 5/12

Renee has to select such that she selects one candy out of 2 sour candy. Clearly that means that the other candy has to be selected out of 4 sweet candies.
This can be done in 2C1 * 4C1 ways = 2*4 = 8 ways

Total ways to select 2 candies out of 6 candies: 6C2: $$\frac{6!}{4!2!}$$ = $$\frac{6.5}{2}$$ = 15 ways.

Probability = $$\frac{8}{15}$$

Option C is correct
Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of   [#permalink] 28 May 2016, 19:07
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