self doubt

Question

Eric throws two dice, and his score is the sum of the values shown.  Sandra throws one die, and her score is the square of the value shown.  What is the probability that Sandra’s score will be strictly higher than Eric’s score?

a.  137/216

b.  17/36

c.  173/216

d.  5/6

Answer 1

All possible combination= 6*6*6 = 216

If Sandra's dice=1 => her score=1 ==>no winning case

If Sandra's dice=2 => her score=4 ==> wins if Eric dice= (1,1) (1,2) (2,1) ==>3 ways

If Sandra's dice=3 => her score=6 ==> wins if Eric dice= (1,1)..(1,6);(2,1)..(2,6);(3,1)..(3,5);(4,1)..(4,4);(5,1)..(5,3);(6,1)..(6,2)

==>(6+6+5+4+3+2)=26 ways

If Sandra's dice=4 or 5 or 6 => her score=16 or 25 or 36 >12(max Eric's score) ==> 36+36+36=108 ways

thus, total winning cases for Sandra= (3+26+108) =137 ways

thus,  probability that Sandra’s score will be strictly higher than Eric’s score = 137/216

Comments

@Naveen

Yes answer is 137/216

Thanks!

Answer 2

Sorry for that i read question wrongly ( sandra also throws 2 dices )... But for my satisfaction i keep it as it is

DICE = {1,2,3,4,5,6}

Sum possible with rolling Two dices = {2,3,4,5,6,7,8,9,10,11,12}

favorable cases for

SUM = 2 ------ { (1,1) } =====> $\frac{1}{36}$

SUM = 3 ------ { (1,2),(2,1) } =====> $\frac{2}{36}$

SUM = 4 ------ { (1,3),(2,2),(3,1) } =====> $\frac{3}{36}$

SUM = 5 ------ { (1,4),(2,3),(3,2),(4,1) } =====> $\frac{4}{36}$

SUM = 6 ------ { (1,5),(2,4),(3,3),(4,2),(5,1) } =====> $\frac{5}{36}$

SUM = 7 ------ { (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) } =====> $\frac{6}{36}$

SUM = 8 ------ { (2,6),(3,5),(4,4),(5,3),(6,2) } =====> $\frac{5}{36}$

SUM = 9 ------ { (3,6),(4,5),(5,4),(6,3) } =====> $\frac{4}{36}$

SUM = 10 ------ { (4,6),(5,5),(6,4) } =====> $\frac{3}{36}$

SUM = 11 ------ { (5,6),(6,5) } =====> $\frac{2}{36}$

SUM = 12 ------ { (6,6) } =====> $\frac{1}{36}$

 

Eric Sum would be {2,3,4,5,6,7,8,9,10,11,12}

when Eric Sum=2, Sandra's Sum should be grater than 2

when Eric Sum=3, Sandra's Sum should be grater than 3 and etc..

 

Required Probability

= { ( Eric=2 and Sandra >2 ) + ( Eric=3 and Sandra >3 ) + ( Eric=4 and Sandra >4 ) + ...... + ( Eric=12 and Sandra >12 ) }

= $ ( \frac{1}{36}  * \frac{35}{36} )+ ( \frac{2}{36}  * \frac{33}{36} )+ ( \frac{3}{36}  * \frac{30}{36} )+ ( \frac{4}{36}  * \frac{26}{36} )+ ( \frac{5}{36}  * \frac{21}{36} )+ ( \frac{6}{36}  * \frac{15}{36} )+ ( \frac{5}{36}  * \frac{10}{36} )+ ( \frac{4}{36}  * \frac{6}{36} )+ ( \frac{3}{36}  * \frac{3}{36} )+ ( \frac{2}{36}  * \frac{1}{36} )+ ( \frac{1}{36}  * \frac{0}{36} ) $

= $ \frac{1}{36^2} $ . {(35*1)+(33*2)+(30*3)+(26*4)+(21*5)+(15*6)+(10*5)+(6*4)+(3*3)+(2*1)+(1*0)}

= $ \frac{1}{1296} $ . {(35)+(66)+(90)+(104)+(105)+(90)+(50)+(24)+(9)+(2)+(0)}

= $ \frac{1}{1296} $ . {(575)}

= $ \frac{575}{1296} $ 

 

Comments

sandra is rolling one die

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& her score is value of die ..you are solving for if she also throws 2 dice & her score is sum of dice.

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yes @ Naveen.

i updated those information in my answer