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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, 01:21

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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?

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 !!

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..

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

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11 Apr 2011, 01:38

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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 !!

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!

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

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

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07 Nov 2013, 16:05

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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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 !!

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..

\(P(Sour=1)=2*\frac{4}{6}*\frac{2}{5}=\frac{8}{15}\), 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.

\(P(Sour=1)=2*\frac{4}{6}*\frac{2}{5}=\frac{8}{15}\), 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.
_________________

\(P(Sour=1)=2*\frac{4}{6}*\frac{2}{5}=\frac{8}{15}\), 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

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.
_________________

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, 02:51

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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PS Please send me PM if I do not respond to your question within 24 hours.

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, 10:46

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, 18:07

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

gmatclubot

Re: Renee has a bag of 6 candies, 4 of which are sweet and 2 of
[#permalink]
28 May 2016, 18:07

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