Select numbers a and b between 0 and 1 independently and at random, an

Question

Select numbers a and b between 0 and 1 independently and at random, and let c be their sum. Let A, B and C be the results when a, b and c, respectively, are rounded to the nearest integer. What is the probability that A + B = C?

(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 3/4

(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 3/4

deushyant · Answer

We have following possibilities(without order) for A,B,C:
000 √
001 ×
011 √
111 ×
110 ×
101 √
100 ×
010 ×

Hence 3/4 probability for A+B=C

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vasuca10 · Answer

Sir please explain how the answer is 3/4

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chetan2u · Answer

We will have to distribute the cases depending on the values of C.. C can be 0, 1 and 2 Case I :- C=0, which means even A and B should be 0 => A+B=C....0+0=0 so c or a+b<\frac{1}{2} , that is c<\frac{1}{2} Also a<\frac{1}{2} and b<\frac{1}{2} ......Prob = \frac{1}{2}*\frac{1}{2}=\frac{1}{4} But c can vary from 0 to 1 when both a and b are less than 1/2 ...Prob = \frac{1}{2} Overall prob = \frac{1}{4}*\frac{1}{2}=\frac{1}{8} Case II :- C=1, which means only one of A and B should be 1 and other 0 => A+B=C....0+1=1 or 1+0=1 so c or 0.5\frac{1}{2} ......Prob = \frac{1}{2}*\frac{1}{2}=\frac{1}{4} But c can vary from 0.5 to 1.5 in this case. ...Prob = 1 , Overall prob = \frac{1}{4}*1*2=\frac{1}{2} .... multiplication by 2 because two cases => 0+1=1 or 1+0=1 Case III :- C=2, which means both A and B should be 1 => A+B=C....1+1=2 so c or a+b>1.5 , that is c>\frac{3}{2} Also a>\frac{1}{2} and b>\frac{1}{2} ......Prob = \frac{1}{2}*\frac{1}{2}=\frac{1}{4} But c can vary from 1 to 2 when both a and b are greater than 1/2 ...Prob of >1.5 = \frac{1}{2} Overall prob = \frac{1}{4}*\frac{1}{2}=\frac{1}{8} Final probability = \frac{1}{8}+\frac{1}{8}+\frac{1}{2}=\frac{3}{4} E

Lipun · Answer

chetan2u sir,
can you please confirm if the below understanding is correct.

If we round up any number between 0 and 1, then there are only two possible ranges x < 0.5 or x >= 0.5
c can take 3 values - 0,1,2.

The probability of A+B=C can be found using negation technique. We need to check for the probability of A+B != C and subtract that from 1.

The possible scenarios where A+B != C will be:

1. a,b < 0.5 and c >= 0.5 => A=0, B=0, and C=1
For each number there are only two possible ranges and one desired range. So the probability that a and b are less than 0.5 and c is greater than or equal to 0.5 is given by 0.5*0.5*0.5 = 0.5^3 = 1/8

2. a,b >= 0.5 and c < 1.5 => A=1, B=1, and C=1
As is in case 1, the probability of the above result is 0.5*0.5*0.5 = 1/8

Total probability = 2*1/8 = 1/4

P(A+B=C) = 1 - P(A+B!=C) = 1 - 1/4 = 3/4

Ans: E

chetan2u · Answer

You are correct.
You have missed one case although it does not affect the answer. 
a<0.5, b>0.5, and vice-versa, but in these cases c would lie between 0.5 and (1+0.5), so C will always be 1. So A+B=C...0+1=1
Thus, this does not add up to negation as probability is 0 of this not happening.

prateekpsm · Answer

I'm getting an answer of 2/3.

Can someone please tell me what is wrong with my method. Plus  chetan2u  i find your method a bit complicated.

A & B can only attain values as 0 or 1 while C can be 0, 1 or 2. In this also there are conditions like C cannot be 2 even if either of A or B is 0. Hence in my opinion ABC has only 6 possible solutions.

ABC
000
001
011
101
111
112

In above 6 possibilities only 4 Agree with the question and hence the answer should be 4/6 or 2/3.

Please correct me where i'm wrong.

chetan2u · Answer

The method you have taken is wrong. 
C can be 1 when c is between 0.5 and 1.5, so it cannot be the same as C being 2, which is possible only when c is between 1.5 and 2.

You may find method complicated, but then that is the solution as you have to take into account values of a, b and c also, and not just values of A, B and C. 
A, B and C are dependent on a, b and c respectively.

prateekpsm · Answer

chetan2u  Thanks for a quick reply, but i have considered all the possibilities. I will give examples also, just endure with me one last time.

ABC a b c
000 0.2 0.2 0.4
001 0.4 0.2 0.6
011 0.2 0.6 0.8
101 0.6 0.2 0.8
111 0.6 0.6 1.2 
112 0.8 0.8 1.6

Is there any other possibility?

chetan2u · Answer

Possibilities are within each case. 
000 can mean 
0.124, 0.12, 0.244
0.1, 0.3, 0.4
0.3, 0.1, 0.4 ans so on
These possibilities do not get considered in taking 000 as just one case.

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