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
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Question : –
Showing that a single throw with two dice, the chance of throwing
- more than 7 is equal to that of throwing less than 7
- 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