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

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l0rrie
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Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink] New post   Updated on: 08 Nov 2013, 01:23
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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

Show Spoiler
Here's what I did: I thought why not use combinatorics lol (too brave right? :roll: ). 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 !!

Here's the answer:
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! :wink:
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C

Originally posted by l0rrie on 11 Apr 2011, 02:21.
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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fluke
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Re: Rene has a bag of 6 candies, probab& combinatorics possible? [#permalink] New post   11 Apr 2011, 02:38
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l0rrie wrote:
Hi everyone,

I'll be needing some help AGAIN.. Yeah I know sorry :oops: 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? :roll: ). 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 !!

Here's the answer:
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! :wink:


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: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink] New post   08 Nov 2013, 01:24
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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

Show Spoiler
Here's what I did: I thought why not use combinatorics lol (too brave right? :roll: ). 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 !!

Here's the answer:
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! :wink:


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

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.

Answer: C.
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Re: Rene has a bag of 6 candies, probab& combinatorics possible? [#permalink] New post   11 Apr 2011, 03:24
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Thank you so much fluke! :o I totally get it now!

This forum is such an amazing help!
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink] New post   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.

Answer: C.


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] New post   05 Jan 2014, 06:48
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Conquistador22 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.

Answer: C.


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] New post   28 Jun 2014, 06:17
Bunuel wrote:
Conquistador22 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.

Answer: C.


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] New post   28 Jun 2014, 07:06
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tyagigar wrote:
Bunuel wrote:
Conquistador22 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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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink] New post   28 May 2016, 19:07
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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


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
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink] New post   03 Sep 2019, 09:54
Bunuel wrote:
tyagigar wrote:
Bunuel wrote:

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.


I have considered Probability of getting exactly one sour candy as "1 - (Probability of getting 2 sweets) - (Probability of getting 2 sours)"

So,

\(P (exactly 1 sour) = 1 - (\frac{4}{6}*\frac{3}{5}) - (\frac{1}{3}*\frac{1}{5}) = 1 - \frac{1}{5} - \frac{2}{5} = \frac{2}{5}\)

What am I missing/doing wrong?
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink] New post   03 Sep 2019, 10:01
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alexautran wrote:
Bunuel wrote:
tyagigar 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?


I have considered Probability of getting exactly one sour candy as "1 - (Probability of getting 2 sweets) - (Probability of getting 2 sours)"

So,

\(P (exactly 1 sour) = 1 - (\frac{4}{6}*\frac{3}{5}) - (\frac{1}{3}*\frac{1}{5}) = 1 - \frac{1}{5} - \frac{2}{5} = \frac{2}{5}\)

What am I missing/doing wrong?


Check math: 1 - 4/6*3/5 - 2/6*1/5 = 1 - 2/5 - 1/15 = 8/15.
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink] New post   30 Dec 2023, 09:22
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?
Sol=> for the two candies the number of ways to pick a 1st candy which is sour and the other is sweet is (2/6)*(4/5) but the order can be anything 1st sour then sweet or first sweet then sour so they can be arranged in 2! ways so the answer will be(2/6)*(4/5)*2!= 8/15
Hence option C
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink] New post   14 Jan 2024, 00:15
Bunuel wrote:
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

Show Spoiler
Here's what I did: I thought why not use combinatorics lol (too brave right? :roll: ). 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 !!

Here's the answer:
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! :wink:


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

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.

Answer: C.




Simultaneously means the candies are picked at the same time. So why are treating them as if they’re picked one after the other

Posted from my mobile device
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Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of [#permalink] New post   14 Jan 2024, 01:04
Expert Reply
Sazimordecai wrote:
Bunuel wrote:
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

Show Spoiler
Here's what I did: I thought why not use combinatorics lol (too brave right? :roll: ). 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 !!

Here's the answer:
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! :wink:


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

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.

Answer: C.




Simultaneously means the candies are picked at the same time. So why are treating them as if they’re picked one after the other

Posted from my mobile device


Mathematically, the probability of picking items simultaneously, or selecting them one by one 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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