Mathematics GATE 2011 probability

Question

A fair die is tossed two times. the probability that 2nd toss results in value greater than first toss is ?

Answer 1

total outcomes = 36

expected outcomes={(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)}=15

hence Probability=15/36=5/12

Comments

Pls explain me how are you writing this sample space

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as it is mentioned in question :  2d toss results in value greater than first

so if you assume that first toss results in outcome as : 2

now in case of tossing the coin 2nd time , allowed outcomes will be only :  (2,3),(2,4),(2,5),(2,6)

here we cannot include (2,1) and (2,2) as  1<2 and 2=2 but we are only allowed to have a greater value in case of 2nd toss! 

hope u will get it!

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Ok. Thx :)
I was confused by ordered pair
(x,y) x<y  x belongs to first throw and y belongs to sexond throw .  , now i get it

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welcome! :)

Answer 2

Answer is {n(n-1)/2}/{n^2} = (n-1)/(2*n)

Here n = 6 so answer is 5/12.

Comments

{n(n-1)/2} .... for what ?? i am nt getting it ... r u considering (5,6) as 1 ... (4,5) ,(4,6) as 2 ... ??

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given answer is absolutely wrong!

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This is how the formula is derived,

On the first die is rolled getting an outcome is 1/6

On the second roll we need a value greater than the value appeared on the first roll,

ie if the first outcome was 1, we can have 5 values greater than 1, so this probability is 5/6.

if the first outcome was 2, we can have 4 values greater than 2 in second roll, so this probability is 4/6

and so on we get 3/6 for 3, 2/6 for 4, 1/6 for 5.

So the required probability is given by, [1/6]*[(5/6)+(4/6)+(3/6)+(2/6)+(1/6)] = [1/(6^2)]*[1+2+3+4+5]

which is equivalent to above expression. The answer 15/36 = 5/12

Answer 3

..basic sample space analysis