What is the probability of getting a number on the first throw greater

Question

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)  \frac{1}{6}

(B)  \frac{2}{3}

(C)  \frac{5}{12}

(D)  \frac{7}{12}

(E)  \frac{5}{6}

Jayeshvekariya · Accepted Answer

A= number on first throw
B= number on 2nd throw

There are 3 options here.
A=B, A<B, A>B.

The probability of A=B is 1/6.
So the probability of A<B and A>B is 5/6.
1/2 of which be greater and 1/2 will be lower.
So 1*5/2*6 =5/12 is the probability for A>B.
Ans C.

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chetan2u · Answer

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

mvictor · Answer

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

mbaprep2016 · Answer

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

rishit1080 · Answer

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

VampSinatra · Answer

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

danbrown9445 · Answer

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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What is the probability of getting a number on the first throw greater

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