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Three items are removed from a box of 12 items. All three items are in

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Bunuel
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Three items are removed from a box of 12 items. All three items are in [#permalink] Three items are removed from a box of 12 items. All three items are in   Updated on: 24 Jan 2021, 07:08
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62% (02:55) correct 38% (03:06) wrong based on 78 sessions
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Three items are removed from a box of 12 items. All three items are inspected. The box is rejected if more than one item is found to be faulty. If there are three faulty items in the box, what is the probability that the box is accepted?

A. 7/55
B. 16/55
C. 17/55
D. 39/55
E. 48/55
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E

Originally posted by Bunuel on 16 Jan 2021, 10:04.
Last edited by Bunuel on 24 Jan 2021, 07:08, edited 2 times in total.
Edited the OA.
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chetan2u
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Re: Three items are removed from a box of 12 items. All three items are in [#permalink] Three items are removed from a box of 12 items. All three items are in   16 Jan 2021, 21:56
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Bunuel wrote:
Three items are removed from a box of 12 items. All three items are inspected. The box is rejected if more than one item is found to be faulty. If there are three faulty items in the box, what is the probability that the box is accepted?

A. 7/55
B. 16/55
C. 17/55
D. 39/55
E. 48/55



Total items: 12 and faulty : 3

Two cases that box is accepted
1) none is faulty
P of nF-nF-nF = \(\frac{9}{12}*\frac{8}{11}*\frac{7}{10}=\frac{21}{55}\)
2) one is faulty
P of F-nF-nF = \(\frac{3}{12}*\frac{9}{11}*\frac{8}{10}=\frac{9}{55}\)
Now faulty can be at any one draw, so \(3*\frac{9}{55}=\frac{27}{55}\)

Final P = \(\frac{21}{55}+\frac{27}{55}=\frac{48}{55}\)

E
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RashmikaAnegama
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Re: Three items are removed from a box of 12 items. All three items are in [#permalink] Three items are removed from a box of 12 items. All three items are in   24 Jan 2021, 04:34
Can someone provide the explanation please? or can we have an official explanation?
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sahilvermani
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Re: Three items are removed from a box of 12 items. All three items are in [#permalink] Three items are removed from a box of 12 items. All three items are in   24 Jan 2021, 05:08
1st is faulty 2nd is faulty: Probability = 3/12*2/11
1st is faulty 3rd is faulty: Probability = 3/12*9/11*2/10
2nd is faulty 3rd is faulty: Probability = 9/12*3/11*2/10

So, probability that the box is rejected = 3/12*2/11 + 3/12*9/11*2/10 + 9/12*3/11*2/10 = (60 + 54 + 54)/(12*11*10) = 7/55

So, probability that the box is accepted = 1 - 7/55 = 48/55

So, answer should be E. But it's marked as D. What am I missing. Please help.
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chetan2u
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Re: Three items are removed from a box of 12 items. All three items are in [#permalink] Three items are removed from a box of 12 items. All three items are in   24 Jan 2021, 05:21
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RashmikaAnegama wrote:
Can someone provide the explanation please? or can we have an official explanation?


The answer is E as also shown in above posts.
OA is wrongly marked as D.
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ruis
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Three items are removed from a box of 12 items. All three items are in [#permalink] Three items are removed from a box of 12 items. All three items are in   03 Dec 2023, 10:18
Is there a combinatorics answer to this question?
Could it be C (3,3) * C (3,2) / C (12,3)? So, desired outcome (all not faulty * just 1 faulty) / all possible outcomes
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Three items are removed from a box of 12 items. All three items are in [#permalink] Three items are removed from a box of 12 items. All three items are in   03 Dec 2023, 13:12
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ruis wrote:
Is there a combinatorics answer to this question?
Could it be C (3,3) * C (3,2) / C (12,3)? So, desired outcome (all not faulty * just 1 faulty) / all possible outcomes


Not quite.

All not faulty is selecting three from the nine good ones:

C(9,3)= 84

Just one faulty is selecting 1 from the 3 faulty and 2 from the 9 good ones, or:

C(3,1)*C(9,2) = 108

These cases are added, not multiplied. Multiplication implies that each case is paired with the others. This is not true since 1 faulty and none faulty are independent:


84+108= 192

Divided by C(12,3)= 220

192/220=96/110=48/55

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Re: Three items are removed from a box of 12 items. All three items are in [#permalink] Three items are removed from a box of 12 items. All three items are in   03 Dec 2023, 19:58
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I like to solve probabilities as total no. in which event occurs/ total events possible

Total events possible = 12C3 (as we choose 3 items out of 12) = 220

Prob that box is accepted , will include either all items working or 1 item faulty.

All items working = 9C3 (as out of the 9 working items we choose 3) = 84

One item faulty = 9C2 * 3C1 (as we choose 2 working and 1 faulty out of 9 and 3 items respectively) = 108

With the above we get the answer, 192/220 = 43/55

To check the answer further and make sure there are no calculation mistakes , we solve the entire question and sum up the total no of individual events possible. This sum should match with total no. of events.

Two faulty = 9C1 * 3C2 = 27

All faulty = 3C3 = 1

Total cases = 192+27+1= 220

Hence we can be certain that there are no calculation mistakes.

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Three items are removed from a box of 12 items. All three items are in [#permalink] Three items are removed from a box of 12 items. All three items are in   04 Dec 2023, 09:01
Required probability.

Case 1

One faulty

3C1 X 9C2/12C3

3 x 36/220

108/220


Case 2

None faulty

9C3/12C3

84/220

Therefore

108/220 + 84/220

192/220

48/55

Answer E
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Three items are removed from a box of 12 items. All three items are in [#permalink] Three items are removed from a box of 12 items. All three items are in   05 Dec 2023, 15:42
Bunuel wrote:
Three items are removed from a box of 12 items. All three items are inspected. The box is rejected if more than one item is found to be faulty. If there are three faulty items in the box, what is the probability that the box is accepted?

A. 7/55
B. 16/55
C. 17/55
D. 39/55
E. 48/55


Total cases = \( (12_{C_3})=220 \)

# of ways box can be rejected \(= 3_{c _2}* 9_{c_1}+3_{c_3} = 27+1 =28\)

\(3_{c_2} *9_{c_1} \): Two faulty and one non- faulty.

\(3_{c_3}: \) All \(3\) faulty

Thus favorable cases \(= 220-28 =192\)

Prob \(= \frac{192}{220} = \frac{48}{55} \)

Hope it helps.
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Three items are removed from a box of 12 items. All three items are in [#permalink]
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