In a single throw of 2 dice, find the probability of getting a) a doublet of even numbers b) a total of 7 Most useful answer will be marked brainliest
Answers
Given: Two dice thrown once.
To find: find the probability of getting a) a doublet of even numbers b) a total of 7
Solution:
- Now we have given that two dice thrown once.
- Now we know that:
probability = number of favorable outcomes / total number of possible outcomes
- Now when the dice is thrown, possible outcomes are:
{(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)}
So the total number of possible outcomes are: 36
a) a doublet of even numbers :
Favorable outcomes = (2, 2) , (4, 4) , (6, 6) = 3
So Probability = 3 / 36 = 1 / 12
b) a total of 7 :
Favorable outcomes = (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6
So Probability = 6 / 36 = 1/6
Answer:
So the probability of getting a doublet of even numbers is 1/12 and a total of 7 is 1/6.
Answer:
Total number of exhaustive cases in a single throw of two dice=6×6=36
outcomes are(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Number of doublets of even numbers=3
∴ Required probability= 3/36
= 1/12
total for 7
(1,6);(2,5);(3,4);(4,3);(5,2);(6,1)
no. favourable outcomes = 6
total no. of outcomes = 36
therefore the required probabilty is 6/36
i.e 1/6
.