There are 4 blue disks, 6 green disk and nothing else in the container

Question

There are 4 blue disks, 6 green disk and nothing else in the container. If 4 disk are to be selected one after the another at random and without replacement from the container. What is the probability that 2 selected are blue and 2 selected are green?

A) \(\frac{1}{7}\)

B) \(\frac{5}{14}\)

C) \(\frac{3}{7}\)

D) \(\frac{25}{56}\)

E) \(\frac{12}{25}\)

A)  \frac{1}{7}

B)  \frac{5}{14}

C)  \frac{3}{7}

D)  \frac{25}{56}

E)  \frac{12}{25}

alldaytestprep · Answer

There are 4 blue disks, 6 green disk and nothing else in the container. If 4 disk are to be selected one after the another at random and without replacement from the container. What is the probability that 2 selected are blue and 2 selected are green?

A) \(\frac{1}{7}\)

B) \(\frac{5}{14}\)

C) \(\frac{3}{7}\)

D) \(\frac{25}{56}\)

E) \(\frac{12}{25}\)

The pool of 10 disks we have to choose from is BBBB GGGGGG

Probability of selecting 2 blue and 2 green =  \frac{total combos 2 blue * total combos 2 green}{total possible 4 team combos}

total combos 2 blue = 4C2 = 6

total combos 2 green = 6C2 = 15

total possible 4 team combos = 10C4 = 210

P(2 B & 2 G) =  \frac{(6*15)}{(210)}  =  \frac{90}{210}  =  \frac{3}{7}

Answer: C

Funsho84 · Answer

I think the answer is B

Blue = 4
Green = 6

BBGG- this can be arranged in 5 different ways: BBGG,BGGB, BGBG, GGBB, GBGB

BBGG
Now 1st B = 4/10
2nd B = 3/9
1st G = 6/8
2nd G = 5/7

(4/10)* (3/9)*(6/8)*(5/7) = 1/14

Now since this can be arranged in 5 different ways, you have to multiply by 5.

5*(1/14) = 5/14

Answer B

alldaytestprep · Answer

You forgot GBBG. Total is 6/14 = 3/7.

Answer: C

EgmatQuantExpert · Answer

Solution

Given:  
• There are 4 blue disks and 6 green disks in a container, and nothing else is present in the container
• 4 disks are to be selected one after the other at random without replacement

To find:  
• What is the probability that out of the selected 4 disks, 2 are blue and 2 are green?

Approach and Working:   
• Out of total 10 disks, the number of ways one can select 4 disks =  ^{10}C_4  = 210
• Out of 4 blue disks, the number of ways one can select 2 disks =  ^4C_2  = 6
• Out of 6 green disks, the number of ways one can select 2 disks =  ^6C_2  = 15

Hence, the required probability =  \frac{(6 * 15)}{210}  =  \frac{3}{7}

Hence, the correct answer is option C.

Answer: C

Alternate way: 
Selecting blue =  \frac{4}{10} * \frac{3}{9} 
Selecting green =  \frac{6}{8} * \frac{5}{7} 
Hence, probability =  \frac{4}{10} * \frac{3}{9} * \frac{6}{8} * \frac{5}{7} * \frac{4!}{{2! 2!}}  =  \frac{3}{7}

Funsho84 · Answer

Silly mistake on my part. You are right

Posted from my mobile device

ScottTargetTestPrep · Answer

The number of ways to select 2 blue disks is:

4C2 = (4 x 3)/2! = 6

The number of ways to select 2 green disks is:

6C2 = (6 x 5)/2! = 15

The total number of ways to select 4 disks from 10 is:

10C4 = 10!/(4! x 6!) = (10 x 9 x 8 x 7)/4! = (10 x 9 x 8 x 7)/(4 x 3 x 2 x 1) = 10 x 3 x 7 = 210

So, the total probability (6 x 15)/210 = 15/35 = 3/7.

Answer: C

pikolo2510 · Answer

Hey  Bunuel ,

Can you tell me where I went wrong?

The combinations possible are

BBBB --> 1 Way
BBBG -->  4!/3!  --> 4 ways
BBGG -->  4!/(2!*2!)  --> 6 ways
BGGG -->  4!/3!  --> 4 ways
GGGG --> 1 Way

Total # of combinations = 16 ways

Therefore, probability =  6/16 
= 3/8

Kindly help

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There are 4 blue disks, 6 green disk and nothing else in the container

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