ANALYTICAL THINKI...
Consider a tetrahedral die and roll it twice. What is the probability that the number on
the first roll is strictly higher than the number on the second roll? For tetrahedral dice
having four faces
Answers
Given : a tetrahedral die rolled twice.
To Find : probability that the number on the first roll is strictly higher than the number on the second roll
Solution:
Tetrahedral die has 4 faces
Numbered 1 , 2 , 3 , 4
roll it twice.
Possible outcomes = 4 x 4 = 16
{ ( 1, 1) , ( 1 , 2) , ( 1 , 3) , ( 1 , 4) ,
( 2, 1) , ( 2, 2) , ( 2, 3) , ( 2 , 4) ,
( 3, 1) , ( 3 , 2) , ( 3 , 3) , ( 3 , 4) ,
( 4, 1) , ( 4 , 2) , ( 4 , 3) , ( 4 , 4) }
number on the first roll is strictly higher than the number on the second roll
= { ( 2, 1) , ( 3, 1) , ( 3 , 2) , ( 4, 1) , ( 4 , 2) , ( 4 , 3) }
6 possible out comes
probability that the number on the first roll is strictly higher than the number on the second roll = 6/16
= 3/8
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