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

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

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

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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Conquistador22

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 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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Bunuel

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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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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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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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

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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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

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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