Question

Two dice are rolled simultaneously.Find the probability of getting (a) an odd number both side (b) sim of number as a multiple of 2 (c) sum of numbers as prime number (d) not doublet (e) difference of digit as 2(f) sum of numbers as at least 10

Answer 1

Answer:

Given:-

• two dice are rolled simultaneously

To find:-

1. an odd number both side
2. sum of number as multiple of 2
3. sum of number as a prime number
4. not doublet
5. difference of digit as 2
6. sum of numbers as at least 10

Solution:-

when two dice are rolled simultaneously;

the total number of possible outcome

$= 6 \times 6 = 36$

1) an odd number both side

in this case

( 1 ,1 ) ( 1,3 ) ( 1,5 ) ( 3, 1)( 3,3 ) ( 3,5)(5 , 1)( 5 ,3) ( 5, 5)

the favrable out come = 9

$required \: probability = \frac{9}{36} = \frac{1}{4}$

2 )sum of number as multiple of 2

The favorable outcome are,

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

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

( 6 ,4 ) ( 6 ,6)

number of favourable outcome = 18

$required \: probability = \frac{18}{36} = \frac{1}{2}$

3)sum of number as a prime number:-

The favourable out come are

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

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

number lf favourable outcome = 14

$required \: probability = \frac{14}{36} = \frac{7}{18}$

4)not doublet:-

The favourable out come are ,

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

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

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

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

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

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

Number of favourable outcome = 30

$required \: probability = \frac{30}{36} = \frac{15}{18} = \frac{5}{6 }$

4 ) difference of didit is 2 :-

the favourable out come are,

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

number of favourable out come= 8

$required \: \: probability = \frac{8}{36} = \frac{4}{18} = \frac{2}{9}$

5 ) sum of number at least 10:-

the favourable out come are ,

( 4 ,6) ( 5 ,5) ( 5, 6) ( 6, 4) ( 6,5) ( 6,6)

number of favourable outcome = 6

$required \: probability = \frac{6}{36} = \frac{1}{6}$