Question

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?

Answer 1

We have to find, the probability that the number on the first roll is strictly higher than the number on the second roll.

When tetrahedral die are roll twice.

We know that:

Tetrahedral die has four faces(1, 2, 3 and 4).

All possible outcomes are:

(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).

Total number of possible outcomes = 16

All favourable outcomes are:

(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)

The number of  favourable outcomes = 6

∴ Probability, P(E) = $\dfrac{6}{16}$ or, $\dfrac{3}{8}$

Thus, the requires probability is $\dfrac{6}{16}$ or, $\dfrac{3}{8}$.

Answer 2

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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