Question

(xv) The probability of not getting doublets in a single throw with pair of dice is :
(a)25/36
(b)6/36
(c)1/36
(d) none of these

​

Answer 1

Given : single throw with pair of dice  

To Find : probability of not getting doublets

a)25/36

(b)6/36

(c)1/36

(d) none of these​

Solution:

A dice has 6 possible out comes = { 1 , 2 ,  3 ,4  , 5 ,6 }

pair of dice is thrown

Hence possible outcomes  = 6 x 6 = 36

Doublet { (1 , 1) , ( 2 , 2) , ( 3 , 3) , ( 4 , 4) , ( 5 , 5) , ( 6  6) }

6 possible doublets

not doublets = 36 - 6  =  30

probability of not getting doublets  =  30/36

= 5/6

Hence none of these is correct answer

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Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

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

(a)25/36

(b)6/36

(c)1/36

(d) none of these

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

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

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

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

FINAL ANSWER

Hence the correct option is

(d) None of these

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