Question

In a single throw of two dice ,find probability of getting (a) doublets
(b) a total of 11

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Answer 1

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 2

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