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Difficult Math Question #67 - Combinations

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Answer

by lalitaroral » Wed Nov 29, 2006 10:22 pm
D:Defect, C:Correct

Possible combinations to select 4 TV(condition atleast 2 should be defective)

1. 3D 1C
2. 2D 2C

Case 1: 7C1*3D3
Case 2: 7C2*3D2

Total Possible ways: 7C1*3D3 + 7C2*3D2 = 84 Ways
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Re: Difficult Math Question #67 - Combinations

by kulksnikhil » Fri Dec 01, 2006 3:55 am
800guy wrote:A shipment of 10 TV sets includes 3 that are defective. In how many ways can a hotel purchase 4 of these sets and receive at least two of the defective sets?
Possible ways

1> All 3 defective and 1 correct
2> 2 defective and 2 correct


1> 1 * 7 = 7
2> 3C2 * 7C2 = 6 * 21 = 126

total= 126 + 7 = 133
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OA

by 800guy » Fri Dec 01, 2006 5:26 pm
OA:

There are 10 TV sets; we have to choose 4 at a time. So we can do that by 10C4 ways. We have 7 good TV’s and 3 defective.

Now we have to choose 4 TV sets with at least 2 defective. We can do that by

2 defective 2 good
3 defective 1 good

That stands to 3C2*7C2 + 3C3*7C1 (shows the count)

If they had asked probability for the same question then

3C2*7C2 + 3C3*7C1 / 10C4.
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Re: Difficult Math Question #67 - Combinations

by jasminekaur280 » Wed Jun 03, 2020 12:33 am
There are 10 TV sets; we have to choose 4 at a time. So we can do that by 10C4 ways. We have 7 good TV’s and 3 defective.

Now we have to choose 4 TV sets with at least 2 defective. We can do that by

2 defective 2 good
3 defective 1 good

That stands to 3C2*7C2 + 3C3*7C1 (shows the count)

If they had asked probability for the same question then

3C2*7C2 + 3C3*7C1 / 10C4.
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Re: Difficult Math Question #67 - Combinations

by quichopipe » Wed Jun 03, 2020 12:55 pm
If we let N = normal TV and D = defective TV, there are two ways we can receive at least 2 defective TVs
Case 1: DDNN
Case 2: DDDN
(DDDD can't happen because there are only 3 defective TVs)

Therefore, the total number of ways that the hotel can receive at least 2 defective TVs is

(3C2)*(7C2) + (3C3)(7C1) = 70 ways
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