In a single throw of two dice ,find probability of getting (a) doublets
(b) a total of 11
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Answers
Answer:
a) Probability of getting Doublet is \frac{1}{6}
6
1
b) Probability of getting total of 11 is \frac{1}{18}
18
1
Solution:
Given that, In a single throw of 2 dice
To find: probability of getting a) Doublets b) a total of 11
The probability of an event is given as:
probability =\frac{\text { number of favorable outcomes }}{\text { total number of possible outcomes }}probability=
total number of possible outcomes
number of favorable outcomes
Here in a single throw of 2 dice, total number of possible outcomes is given as:
{(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)}
Total number of possible outcomes = 36
a) probability of getting Doublet
Favorable outcomes = (1, 1) , (2, 2) , 93, 3) , (4, 4) , (5, 5) , (6, 6)
Number of favorable outcomes = 6
\text{ probability of getting doublet } = \frac{6}{36} = \frac{1}{6} probability of getting doublet =
36
6
=
6
1
b) probability of getting total of 11
Favorable outcomes = (5, 6) , (6, 5)
Number of favorable outcomes = 2
\text{ probability of getting total of 11 } = \frac{2}{36} = \frac{1}{18} probability of getting total of 11 =
36
2
=
18
1
Learn more about probability
In a single throw of two dice what is the probability of getting sum of 9
https://brainly.in/question/2464166
In a throw of two dice,find the probability of getting a sum of 10
Answer:
(a) 1/6
(b) 1/18
Step-by-step explanation:
Probability = No. of Required output
No. of Total output
Total output = { (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) } = 36
( a ).Double
Required output = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) } = 6
Probability = 6/36
=> Probability = 1/6
( b ) Total of 11
Required output = { (5,6), (6,5) } = 2
Probability = 2/36
=> Probability = 1/18