Latest TCS Aptitude Question SOLUTION: Eric throws two dice nd his score is the sum of the values shown. sandra throws one dice nd her score is the square of the value shown. what is the probability that sandra's score

Ans: 137/216 (STAR QUESTION)

Thought process:
At first we must seek the conditions at which SANDRA strictly wins.
As Sandra's score is always the square of the dice value:
Dice Value Sandra's Score
1 1
2 4
3 9
4 16
5 25
6 36
So, from dice value 4 on wards we are absolutely sure that Sandra would win (As Eric's score is sum of two dice which would never exceed 6+6 =12)
So, we can conclude that:
Probability of SANDRA winning is 1 if she gets 4,5 or 6.
Now we need to consider as to what will happen if Sandra got a 1,2 or a 3?
Probability of SANDRA winning is 0 if she gets a 1 (As in that case Eric would score a minimum of 1+1=2)
What will happen if SANDRA gets a 2 or a 3?

SANDRA gets '2' (means she scored 4)
---------------
Favorable events are if ERIK gets (1,1),(1,2) or (2,1) [Find all combinations which is less than 4]
Total no of favorable events= 3
Probability= 3/36 where 36 is coz total 36 events are possible

SANDRA gets a '3' (means she scored 9)
-----------------
She wins if Erik does not get (6,6),(6,5)(5,6),(6,4),(4,6)(5,5),(6,3),(3,6),(4,5),(5,4)........... Total= 36-10= 26 favorable events. [Here we just went by the compliment approach for simplicity!]
Therefore probability = 26/36.

Probability of Sandra getting any of the 6 nos is 1/6.
HENCE, PROBABILITY THAT SANDRA WINS :
=(1/6 * 0)+(1/6 * 3/36)+(1/6 * 26/36)+(1/6 * 1)+(1/6 * 1)+(1/6 * 1)= 137/216

as sample space is 216 i.e total no. of cases
now,
sandra's score no. of favourable cases to win
1 0
4 3 @(1,1)(2,1)(1,2)
9 26 cases all pairs till 9
16 36(as max sum is 12,n 16>12)
25 36
36 36
total favourable cases= 0+3+26+36+36+36= 137
thus,
probablity=137/216

The answer is 137/216

favorable cases are 3 i.e. 4 , 5 , 6 whose square is greater than 6+6
and total cases are 36 so probablity is 3/36 or 1/12