Adam bought a box of 15 Apples out of which exactly 5 were Green and

Question

Adam bought a box of 15 Apples out of which exactly 5 were Green and rest were Red. He took out 4 Apples at random. What is the probability that atleast 2 out of 4 Apples selected were Green?

A. 3/91
B. 13/91
C. 30/91
D. 31/91
E. 37/91

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PyjamaScientist · Answer

Rotated the image posted earlier.

thkkrpratik · Answer

Bunuel  when will the OA be shared?

PyjamaScientist · Answer

OA has been revealed already.

thkkrpratik · Answer

Could you please explain the reason for taking 3x2 in step 1 and 4!/3! in step 2? I do remember it being related to the number of ways in which apples can be selected but cannot recollect the entire thing. Would appreciate your help.

PyjamaScientist · Answer

Hi  thkkrpratik ,

Thank you for your query and your PM. I also just realised that the solution I posted earlier was 90 degrees rotated. I had posted it through my iPhone and thus didn't realise the error. I have rotated the image and the solution is much more easily readable now.

Coming to your query,  3*2  in step 1 is actually:  \frac{4!}{2!*2!}  which upon simplification gives  \frac{4*3*2*1}{2*2} = 3*2 
You need  4!  to calculate the total different arrangements of all four apples chosen in the first step. But since two apples are red and two apples are green, we have to eliminate the repetitions. And thus, the division by  2!  (for two red apples)  * 2!  (for two green apples) comes into the picture.

For the same reason, Since there are three green and one red apple in step two, you have to do  4!  to incorporate all possible arrangements and then divide by  3!  to eliminate repeat arrangements from  3   identical green apples.

Hope it helps.

thkkrpratik · Answer

Understood now, thank you.

HarshavardhanR · Answer

P(At least 2 out of 4 green apples) = 1 -

P(0 green apples) = 10C4/15C4
P(exactly 1 green apple) = (5C1 x 10C3) / 15C4

= 10C4/15C4 + (5C1 x 10C3) / 15C4 = 54/91

Therefore,
P(At least 2 out of 4 green apples) = 1 -

= 1 - 54/91 = 37/91.  Choice E .

---
Harsha

dec2023 · Answer

# of ways to select 4 apples out of 15 => 15C4 =15x7x13

# of ways to select only 1 green apple => 1G3R => 5C1 x 10C3 =5x10x3x4

# of ways to select 0 green apple => 4R => 10C4=10x3x7

P(<1 G) =   / (15x7x13) = 54/91

P(>=1G) = 1 - (54/91) =37/91

Adam bought a box of 15 Apples out of which exactly 5 were Green and

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Adam bought a box of 15 Apples out of which exactly 5 were Green and

Prep Toolkit

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