Question

the probability of not getting diublets in a single throw with pair of dice is​

Answer 1

Step-by-step explanation:

So, the number of favourable outcomes is 3. We know that the number of possible outcomes is 36. Thus, the probability of getting a doublet of an even number is =336=112.

Answer 2

SOLUTION

TO DETERMINE

The probability of not getting doublet in a single throw with pair of dice

EVALUATION

Here it is given that a pair of dice has thrown once

So the event points 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 = 36

Let A be the event that of getting doublet

Then the event points for the event A are

(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)

So the total number of event points for the event A is 6

Here the probability of getting doublet

$\sf{ =P(A) }$

$\displaystyle \sf{ = \frac{6}{36} }$

$\displaystyle \sf{ = \frac{1}{6} }$

Hence the probability of not getting doublet

$\sf{ =P( \overline{A}) }$

$\displaystyle \sf{ =1 - \frac{1}{6} }$

$\displaystyle \sf{ = \frac{5}{6} }$

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