Latest TCS Aptitude Question SOLUTION: There are 2 persons A and B. A select 2 numbers from the set { 3,4,5,6,7}. There are 2 sets { 5,7,19,23,34} and {13,24,4,56,11}. B selects 1 no. from any one of the set . What is

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(#M40028873) TCS question Keep an EYE

There are 2 persons A and B. A select 2 numbers from the set { 3,4,5,6,7}. There are 2 sets { 5,7,19,23,34} and {13,24,4,56,11}. B selects 1 no. from any one of the set . What is the probability that the no. chosen by B is greater than the sum of the 2 numbers selected by A.

Asked In TCS Ankit Bansal (10 years ago)

Unsolved Read Solution (19)

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Siddhartha  (10 years ago)

There can be 5C2=10 combinations when 2 numbers are selected from A.
The sum of the 2 numbers are between : 7 to 13.
so,we have to consider each case from 7 to 13 when 2 numbers are selected.

When sum=7 : probability of selecting sum 7 from A = 1/10 (obtained when {3,4} is selected).After that,probability of selecting a set from B us 1/2.
And when sum 7 we need to select (19,23,34) from set 1 and (13,24,56,11) from set 2. So, total probab = 1/10*[(1/2*3/5)+(1/2*4/5)] = 7/100 .

Similarly,we need to find the cases for the rest of the sums 8 to 13.

Total 65/100 will come rounded to 13/20(Ans).

Is this solution Helpfull? Yes (14) | No   

kirti srivastav  (10 years ago)

ans is 1/2

Is this solution Helpfull? Yes (9) | No (2)  

roushan kumar  (10 years ago)

ans is 59/100...

Is this solution Helpfull? Yes (2) | No (4)  

ronak roy  (10 years ago)

a select number it's sum is atmost=13
so b select number greater then 13
hence
5c2/(5c3+5c2)=10/20=1/2
hence ans=1/2;
is it right??

Is this solution Helpfull? Yes (2) | No (1)  

roushan kumar  (10 years ago)

total 65 cases .... so it should 65/100 i.e. 13/20

Is this solution Helpfull? Yes (2) | No   

chandan kumar  (10 years ago)

A can select {3,4},{3,5},{3,6},{3,7},{4,5},{4,6},{4,7},{5,6},{5,7},{6,7} total 10 ways.
B can select from {5,7,19,23,34,13,24,4,56,11}
then B select greater than 7
then probablity is 7/11

Is this solution Helpfull? Yes (2) | No (1)  

Ravi Kumar Mishra  (10 years ago)

ANS IS 5/10 =1/2;A A SELECTS SET OF NUMBER OF SUM IN SUCH WAY,(7,9,11,13,10,11,12,9,10,8),REMOVING THESE FROM SET OF B SELECTION.

Is this solution Helpfull? Yes (2) | No   

Maithreyi  (10 years ago)

ans is 41/50

Is this solution Helpfull? Yes (1) | No (8)  

Sayan Banerjee  (10 years ago)

Pls explain the solution of this problem.

Is this solution Helpfull? Yes (1) | No (3)  

Triloki Shaw  (10 years ago)

sorry, there is mistake in my previous approach.

Is this solution Helpfull? Yes (1) | No   

sachin  (10 years ago)

ans is 13/20

Is this solution Helpfull? Yes (1) | No (1)  

Dipannita Adhikary  (10 years ago)

ans is 13/20

Is this solution Helpfull? Yes (1) | No   

Shushmita Parashar  (10 years ago)

A can select a no in 5c2 ways.
B has to choose a set a then select a no which is greater than the sum of those two no alredy chosen by A(which could max be 13).Therefore, 1/2*3c1(19,23,24 from one set)+2c1(56,24) ways.
hence req probability=(1/2*(3c1+2c1))/5c2=1/4.

Is this solution Helpfull? Yes (1) | No   

ankit  (10 years ago)

correct ans is 9/20
procedure
we can first select the numbers in 100 ways so
sample space=100
now the sum of two numbers given by A will be
either 7 or 8 or 9 or 10 or 11 or 12 or 13
when sum is 7 then we can select any 7 numbers for B
when sum is 8 we can select again 7 numbers for B
when sum is 9 .................... 7 ............
when sum is 10 ....................7.............
when sum is 11 ...................6.............
when sum is 12 ...................6.............
when sum is 13....................5.............
so total =45
hence probablity =45/100
====9/20

Is this solution Helpfull? Yes (1) | No (2)  

Rahul Singh  (10 years ago)

please explain the procedure

Is this solution Helpfull? Yes | No (1)  

Triloki Shaw  (10 years ago)

A can select two numbers as
i. 3+4=7; ii. 3+5=8; iii. 3+6=9; iv. 3+7=10;
v. 4+5=9; vi. 4+6=10; vii. 4+7=11;
viii. 5+6=11; ix. 5+7=12; 6+7=13
So probability of case i,ii,iii,iv,v,vi is like selecting one number from
19,23.,34,13,24,56,11 i.e., 1/7 for each case and for all 6cases 6/7
Probability of case vii,viii,ix is one from 19,23,34,13,24,56 i.e., 1/6 for each
and 3/6 fo all three.
prob. of last case is 1/5
therefore 6/7+3/6+1/5=327/210= 109/70 ans.

Is this solution Helpfull? Yes | No (15)  

DEEPANSHI SRIVASTAVA  (10 years ago)

5C1/10C1=1/2

Is this solution Helpfull? Yes | No (1)  

puja kundu  (10 years ago)

A can choose 2 numbers out of 5 numbers in 5C2 ways.in such cases sum of 2 numbers is between 7 to 13.now B can choose 5 which is less than 7 and the probability is 1/10.
so,probability that number choosen by B is greater than sum of the numbers of A is 1-1/10=9/10.

Is this solution Helpfull? Yes | No (2)  

Chirag Ranjan Lahiri  (9 years ago)

A can select {3,4},{3,5},{3,6},{3,7},{4,5},{4,6},{4,7},{5,6},{5,7},{6,7} total 10 ways.
B can select from {5,7,19,23,34,13,24,4,56,11}
then B select greater than 7
then probablity is 7/11

Is this solution Helpfull? Yes | No   

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