In a simultaneous throw of a pair of dice, find the probability of get

1.	In a simultaneous throw of a pair of dice, find the probability of getting a sum, which is a perfect square.
Answer» The experiment is throw a pair of dice. Therefore, the sample space is \(S=\begin{Bmatrix}(1,1), (1, 2),{(1,3)},(1,4),(1,5),(1,6)\\(2,1),{(2,2)},(2,3),(2,4),(2,5),(2,6)\\{(3,1)}, (3,2),(3,3),(3,4),(3,5),(3,6)\\(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{Bmatrix}\) Total possible outcomes = n(S) = 6 \(\times\) 6 = 36. Let E = Event of getting the sum as a perfect square. Therefore, E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)} Number of favorable outcomes to event E = n(E) = 7 Thus, probability of getting the sum as a perfect square is p(E) = \(\frac{n(E)}{n(S)}=\frac{7}{36}\)

In a simultaneous throw of a pair of dice, find the probability of getting a sum, which is a perfect square.

Answer»

The experiment is throw a pair of dice.

Therefore, the sample space is

\(S=\begin{Bmatrix}(1,1), (1, 2),{(1,3)},(1,4),(1,5),(1,6)\\(2,1),{(2,2)},(2,3),(2,4),(2,5),(2,6)\\{(3,1)}, (3,2),(3,3),(3,4),(3,5),(3,6)\\(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{Bmatrix}\)

Total possible outcomes = n(S) = 6 \(\times\) 6 = 36.

Let E = Event of getting the sum as a perfect square.

Therefore, E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}

Number of favorable outcomes to event E = n(E) = 7

Thus, probability of getting the sum as a perfect square is

p(E) = \(\frac{n(E)}{n(S)}=\frac{7}{36}\)

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