Question

In a single throw of 2 dice, find the probability of getting a) Doublets b) a total of 11

Answer 1

Total outcomes = 12
Favorable outcome = 6 ( 1,1, 2,2, 3,3, 4,4, 5,5 ,6,6 )
Probability = 6/12 = 1/2


Total outcomes = 12
favorable outcomes = 0
Probability = 0/12 = 0

Answer 2

a) Probability of getting Doublet is $\frac{1}{6}$

b) Probability of getting total of 11 is $\frac{1}{18}$

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:

$probability =\frac{\text { number of favorable outcomes }}{\text { total number of possible outcomes }}$

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

$\text{ probability of getting doublet } = \frac{6}{36} = \frac{1}{6}$

b) probability of getting total of 11

Favorable outcomes = (5, 6) , (6, 5)

Number of favorable outcomes = 2

$\text{ probability of getting total of 11 } = \frac{2}{36} = \frac{1}{18}$

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