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On a standard die one of the dots is removed at random with each dot e

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On a standard die one of the dots is removed at random with each dot e  [#permalink]

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New post 22 Mar 2019, 01:28
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Question Stats:

12% (02:12) correct 88% (02:12) wrong based on 26 sessions

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Re: On a standard die one of the dots is removed at random with each dot e  [#permalink]

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New post 22 Mar 2019, 02:36
Bunuel wrote:
On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

(A) 5/11
(B) 10/21
(C) 1/2
(D) 11/21
(E) 6/11



The question will havw six cases to consider and two events for probab

1. probab of selecting dot to remove

2. probab of getting odd face

Each time we remove a dot,
no of odd faces and even faces will change .

each case has these two probab multiplied .


and for the final solution,
we add all the cases



attached image has the solution in short

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Re: On a standard die one of the dots is removed at random with each dot e  [#permalink]

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New post 22 Mar 2019, 03:40
Bunuel wrote:
On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

(A) 5/11
(B) 10/21
(C) 1/2
(D) 11/21
(E) 6/11


total dots on die = 1+2+3+4+5+6 = 21
case 1 :total even dots ; 2+4+6 = 12
P of even dot is removed; 12/21 ; 4/7
we will have 4 odd dots; 4/7 * 4/6 ; 8/21
case 2; odd dots; 1+3+5= 9
P = 9/21 ; 3/7
1 odd dot is removed so 2 odd faces left ; P 3/7 * 2/6 = 1/7
total P = 8/21 + 1/7 = 8+3 /21 ; 11/21
IMO D
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Re: On a standard die one of the dots is removed at random with each dot e  [#permalink]

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New post 23 Mar 2019, 07:57
Archit3110 wrote:
Bunuel wrote:
On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

(A) 5/11
(B) 10/21
(C) 1/2
(D) 11/21
(E) 6/11


total dots on die = 1+2+3+4+5+6 = 21
case 1 :total even dots ; 2+4+6 = 12
P of even dot is removed; 12/21 ; 4/7
we will have 4 odd dots; 4/7 * 4/6 ; 8/21
case 2; odd dots; 1+3+5= 9
P = 9/21 ; 3/7
1 odd dot is removed so 2 odd faces left ; P 3/7 * 2/6 = 1/7
total P = 8/21 + 1/7 = 8+3 /21 ; 11/21
IMO D
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Re: On a standard die one of the dots is removed at random with each dot e  [#permalink]

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New post 23 Mar 2019, 23:13
Bunuel wrote:
On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

(A) 5/11
(B) 10/21
(C) 1/2
(D) 11/21
(E) 6/11



please share OA. I am unable to get how to go about it.
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Re: On a standard die one of the dots is removed at random with each dot e  [#permalink]

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New post 24 Mar 2019, 03:03
@
Archit3110 wrote:
Bunuel wrote:
On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

(A) 5/11
(B) 10/21
(C) 1/2
(D) 11/21
(E) 6/11


total dots on die = 1+2+3+4+5+6 = 21
case 1 :total even dots ; 2+4+6 = 12
P of even dot is removed; 12/21 ; 4/7
we will have 4 odd dots; 4/7 * 4/6 ; 8/21
case 2; odd dots; 1+3+5= 9
P = 9/21 ; 3/7
1 odd dot is removed so 2 odd faces left ; P 3/7 * 2/6 = 1/7
total P = 8/21 + 1/7 = 8+3 /21 ; 11/21
IMO D


Can you explain, on why 4/7 is multiplied by 4/6 "in case 1" ?

and also if we remove 1 odd dot, we will have 2 evens then, as
1--will become-> 0..,
3--will become-> 2,
and
5--will become-> 4 ...

Is that why, you took 2/6 as the probability "in case 2" ?
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Re: On a standard die one of the dots is removed at random with each dot e  [#permalink]

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New post 24 Mar 2019, 03:47
Hi guys , can you let me know why am I wrong :
we can remove the dot from an even number or an odd number
since we have 3 odd and 3 even numbers , the probability of removing it from an odd number will be 1\2 , same for an even number
now we have 2 scenarios , if we remove it from an even face : we are left with 4 odd faces and 2 even
(and vice versa if we choose an odd face )
so p is 1/2 * 4/6 + 1/2 * 2/6 = 1/2
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On a standard die one of the dots is removed at random with each dot e  [#permalink]

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New post 24 Mar 2019, 04:55
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foryearss wrote:
Hi guys , can you let me know why am I wrong :
we can remove the dot from an even number or an odd number
since we have 3 odd and 3 even numbers , the probability of removing it from an odd number will be 1\2 , same for an even number
now we have 2 scenarios , if we remove it from an even face : we are left with 4 odd faces and 2 even
(and vice versa if we choose an odd face )
so p is 1/2 * 4/6 + 1/2 * 2/6 = 1/2



The red part I highlighted is your mistake
The question didn't ask you to remove an odd or even number (so the probability can't be 1/2), instead it required you to remove a dot. This dot is maybe in any odd face or even face, and the probability of removing this dot for each case is totally different.
- To remove a dot in any even face, the probability is 12/21
- To remove a dot in any odd face, the probability is 9/21

Hope it helps
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On a standard die one of the dots is removed at random with each dot e   [#permalink] 24 Mar 2019, 04:55
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