Probability of More than 7, Less than 7 and Exactly 7 a single throw two dice

In this article explain the statistics topic probability. Her we find or calculate the probability Showing that a single throw with two dice, the chance of throwing more than 7 is equal to that of throwing less than 7 and hence find probability of throwing exactly 7

Question : –

Showing that a single throw with two dice, the chance of throwing

1. more than 7 is equal to that of throwing less than 7
2. probability of throwing exactly 7

Sample space of Two Dice Roll

n(S) = 36

(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)

Probability More Than 7

Let A represent the event that a total of more than 7 occurs. Then the event A has 15 following outcomes

(2,6) (3,5) (3,6) (4,4) (4,5) (4,6) (5,3) (5,4) (5,5) (5,6) (6,2) (6,3) (6,4) (6,5) (6,6)

P(A) = n(A)/n(S) = 15 / 36

Probability Less Than 7

Let B represent the event that a total of less than 7 occurs. Then the event B has 15 following outcomes

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

P(B) = n(B)/n(S) = 15 / 36

Probability Exactly 7

Let C represent the event that a total of exactly 7 occurs. Then the event C has 6 following outcomes

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

P(C) = n(C)/n(S) = 6 / 36

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