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Suppose that a balanced die is rolled three times, and let\({X_i}\)denote the number that appears on the ith roll (i = 1, 2, 3). Evaluate\({\rm P}\left( {{X_1} > {X_2} > X3} \right)\).

Short Answer

Expert verified

Required probability is: \(P\left( {{X_1} > {X_2} > {X_3}} \right) = \frac{5}{{54}}\).

Step by step solution

01

Given information

A balanced die is rolled three times.

Let\({X_i}\)be the number that appear on the ith roll

02

Calculate the value of \({\rm P}\left( {{X_1} > {X_2} > X3} \right)\)

The event when\(\left( {{X_1},{X_2},{X_3}} \right)\)has the following values. is

{(6,5,1), (6,4,1), (6,3,1), (5,4,1), (5,3,1), (4,3,1),(6,5,2), (6,4,1), (6,3,1), (5,4,2), (5,3,2), (4,3,2),(6,5,3), (6,4,3), (6,2,1), (5,4,3), (5,2,1), (4,2,1), (6,5,4), (3,2,1)}

No of elements in the event are 20

Each of these 20 points has the probability\(\frac{1}{{{6^3}}}\)

Hence,

\(\begin{array}{c}P\left( {{X_1} > {X_2} > {X_3}} \right) = \frac{{20}}{{216}}\\ = \frac{5}{{54}}\end{array}\)

Therefore, required probability is: \(P\left( {{X_1} > {X_2} > {X_3}} \right) = \frac{5}{{54}}\).

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