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

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?

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

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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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

Check math: 1 - 4/6*3/5 - 2/6*1/5 = 1 - 2/5 - 1/15 = 8/15.
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Thank you so much fluke! I totally get it now!

This forum is such an amazing help!

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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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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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?

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

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

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Mathematically, the probability of picking items simultaneously, or selecting them one by one without replacement, is the same.
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