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ssr300
What is the probability of getting a number on the first throw greater than that on the second throw when a dice is thrown twice?

A. 1/6
B. 2/3
C. 5/12
D. 7/12
E. 5/6


Hi,

1) simple logic should tell us that answer should be near but LESS than 1/2..
As we are dealing with duplicate patterns.. 1 and 6..... 2 and 5...3 and 4...
C should be the guess..
Why it should be near half is in the method below.

2) there can be six numbers in first throw and six in second , so ways = 6*6*=36..

Now if 1 is in first throw, any number in 2nd throw except 1 itself will give you ans YES, so 5 throw..
But in 2, all except 1 and itself will bE so 6-2=4 ways..
Similarly 3,2,1 and 0 for number 4,3,2,and 1respectively in first throw..
Possible ways=5+4+3+2+1+0=15..

Prob =15/36=5/12..
C

exactly my approach...yet I used approximations.
the odds should be slightly less than 1/2
A is too little
B is too big
C is close
D is too big
E is too big.

C remains
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if in first through prob of getting 1 is 1/6 but in this case other number greater than 1 will be 5/6
similarly prob of getting 2 is 1/6 but in this case other number greater than 2 will be 4/6

adding till 5
1/6( 5/6+ 4/6....1/6)= 5/12

C is the answer
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Probability of getting it is 1/2 minus the probability of getting the same number on both throws which is 1/12.

(1/2) - (1/12) = 5/12

Option C
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total number of outcomes, 6 +6 (1st and 2nd throw )
outcomes on 1st throw==================================123456
in order for outcome to be greater than 1st throw 2nd throw outcome =====23456

so 5 outcomes of higher value ,
answer 5/12.
option c
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ssr300
What is the probability of getting a number on the first throw greater than that on the second throw when a dice is thrown twice?

A. 1/6
B. 2/3
C. 5/12
D. 7/12
E. 5/6

Easy to understand method:

Numbers in brackets are numbers rolled in 2nd throw.

First throw equals 1:
Favorable case when 1st throw > 2nd throw is (not possible) = 0.
Unfavorable case when 1st throw <= 2nd throw is (1,2,3,4,5,6). = 6

First throw equal 2:
Favorable case when 1st throw > 2nd throw is (1) = 1
Unfavorable case when 1st throw <= 2nd throw is (2,3,4,5,6) = 5

First throw equal 3:
Favorable case when 1st throw > 2nd throw is (1,2) = 2
Unfavorable case when 1st throw <= 2nd throw is (3,4,5,6) = 4

First throw equal 4:
Favorable case when 1st throw > 2nd throw is (1,2,3) = 3
Unfavorable case when 1st throw <= 2nd throw is (4,5,6) = 3

First throw equal 5:
Favorable case when 1st throw > 2nd throw is (1,2,3,4). = 4
Unfavorable case when 1st throw <= 2nd throw is (5,6) = 2

First throw equal 6:
Favorable case when 1st throw > 2nd throw is (1,2,3,4,5) = 5
Unfavorable case when 1st throw <= 2nd throw is (6) = 1

Probabality = Total Favorable/Total favorable + unfavorable
=15/36
=5/12
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