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Difficulty:
15%
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Question Stats:
81% (01:45) correct
19%
(02:04)
wrong
based on 1111
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History
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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
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? 
). 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!
). 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!
08 Nov 2013, 01:24
Kudos
Bookmarks
l0rrie
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
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? ). 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!
). 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!
\(P(Sour=1)=\frac{C^1_4*C^1_2}{C^2_6}=\frac{8}{15}\)
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.
General Discussion
11 Apr 2011, 02:38
Kudos
Bookmarks
l0rrie
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!
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
Does that make little sense?
I totally get it now! 
). So I thought there are 6!/4!2! = 15 ways to pick 2 candies out of the 6. 
