Probability - choosing colours/cards

[StoneBlack]

1. Kurt, a painter, has 9 jars of paint:
4 are yellow
2 are red
rest are brown
Kurt will combine 3 jars of paint into a new container to make a new color, which he will name accordingly to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow
Brun X if the paint contains 3 jars of brown paint
Jaune X if the paint contains at least 2 jars of yellow
Jaune Y if the paint contains exactly 1 jar of yellow

What is the probability that the new color will be Jaune

a) 5/42
b) 37/42
c) 1/21
d) 4/9
e) 5/9

I have a question about how to choose the colours after some answers. i would like to contrast this question with another one:

2. In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

Again this involves how you choose the cards.

[Anurag@Gurome]

Kurt, a painter, has 9 jars of paint:
4 are yellow
2 are red
rest are brown
Kurt will combine 3 jars of paint into a new container to make a new color, which he will name accordingly to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow
Brun X if the paint contains 3 jars of brown paint
Jaune X if the paint contains at least 2 jars of yellow
Jaune Y if the paint contains exactly 1 jar of yellow

What is the probability that the new color will be Jaune

a) 5/42
b) 37/42
c) 1/21
d) 4/9
e) 5/9

The probability that the new color will be Jaune is possible in the following situations:
(1) There are at least 2 jars of yellow paint
(2) There is exactly 1 jar of yellow paint

Now let's first find the probability that there are no jars of yellow paint = 5/9 * 4/8 * 3/7 = 5/42

Therefore, probability that some jars are of yellow paint = 1 - 5/42 = 37/42

The correct answer is B.

[StoneBlack]

Thanks Anurag for your reply.

Suppose we do this by choosing colours.

1. 2 yellow means..

a> there can be all three Y => 4c3
OR
b> 2 Y and 1 out of 2 R and 3 B => 4c2 x 5c1

Total 34

2.This has exactly 1 Y and remaining 2 out of 5 = > 4c1 x 5c2

Total 40

Likely possibilities : 34 + 40 = 74
Total possibilities = (9!/3!6!) = 84
Hence probability = 74/84 = 37/42

My question is when choosing for case 1 (Jaune X)
can we not choose in this way : 2 colours of Yellow from 4, 4C2
And Rest one colour can be from the remaining 7 colours left hence, 7C1
or 4C2 * 7C1. Where is the mistake in this?

[saketk]

Thanks Anurag for your reply.

Suppose we do this by choosing colours.

1. 2 yellow means..

a> there can be all three Y => 4c3
OR
b> 2 Y and 1 out of 2 R and 3 B => 4c2 x 5c1

Total 34

2.This has exactly 1 Y and remaining 2 out of 5 = > 4c1 x 5c2

Total 40

Likely possibilities : 34 + 40 = 74
Total possibilities = (9!/3!6!) = 84
Hence probability = 74/84 = 37/42

My question is when choosing for case 1 (Jaune X)
can we not choose in this way : 2 colours of Yellow from 4, 4C2
And Rest one colour can be from the remaining 7 colours left hence, 7C1
or 4C2 * 7C1. Where is the mistake in this?

No, you cannot do that.. you cannot have 7 jars for next selection. Because your case 1 says 2Y + 1 jar of any other color..

So, if you choose 7 then it will also include the 2 Yellow jars. Which is incorrect.

Case 1:- 2Y jars+1 of any other color
case 2:- all 3 are yellow..

I hope this helps.

[saketk]

Answer to your second question--

2. In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

Again this involves how you choose the cards.

Total cards of each type 13

Spade - 13
Diamond - 13
Hearts - 13
Clubs - 13

Clearly, in order to have all types in your selection, you must have 1 of each type first..

So, 1 of each type can be selected in 13 ways each

Rest 2 can be selected out of remaining 48 cards in 48*47 ways..

Therefore, the total comes out to be ..

(13)^4*48*47

Correct Answer -- A.

[navami]

Second question : 13^4 *48*47 ---> choice A