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

Question

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

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

chetan2u · Answer

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

RashmikaAnegama · Answer

Can someone provide the explanation please? or can we have an official explanation?

sahilvermani · Answer

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.

chetan2u · Answer

The answer is E as also shown in above posts. 
OA is wrongly marked as D.

ruis · Answer

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

Regor60 · Answer

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

Posted from my mobile device

karanpatira · Answer

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.

Posted from my mobile device

adewale223 · Answer

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

stne · Answer

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.

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

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