Math, asked by hazelshiny07, 9 days ago

Two dice are thrown simultaneously. Find the probability of getting

a). an even number on first dice

b). an odd number on first dice

c). an even number as the sum

d). a multiple of 5 as the sum

e). a multiple of 7 as the sum

f). a multiple of 3 as the sum

g). a sum more than 7

h). a sum greater than 9

i). neither the sum 9 nor the sum 11
as the sum

j) a sum less than 6

k). a sum less than 7

l). a sum more than 7

m). a multiple of 3 on one dice

n). a multiple of 2 on one dice

o). a multiple of 5 on one dice

p). a multiple of 2 on one dice and a multiple of 3 on the other

q). a doublet

r). a doublet of even number

s). a doublet of odd number

t). a doublet of prime number

u). a number other than 5 on any dice

v). a number other than 3 on any dice

w). the sum equal to 12.

x). the sum than equal to 10

y). the sum less than or equal to 10

z). greater the sum as a prime number

please give correct answerso​

Answers

Answered by tennetiraj86
24

Step-by-step explanation:

Solution :-

Given that

Two dice are thrown simultaneously.

We know that

If n dice are thrown simultaneously then the number of possible outcomes = 6^n

So

If two dice are thrown simultaneously then the total number of possible outcomes = 6^2 = 36

They 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).

a). an even number on first dice :-

Favourable outcomes to even number on the first die =(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

Number of favourable outcomes to the even number on the first die =18

We know that

Probability of an event = P(E)

= Number of favourable outcomes/Total number of possible outcomes

=> Probability of getting an even number on the first die

= 18/36

= 1/2

Probability of getting an even number on the first die = 1/2

b). an odd number on first dice :-

Favourable outcomes to the odd number on the first die = (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

Number of favourable outcomes to the even number on the first die =18

We know that

Probability of an event = P(E)

= Number of favourable outcomes/Total number of possible outcomes

=> Probability of getting an odd number on the first die

= 18/36

= 1/2

Probability of getting an odd number on the first die = 1/2

c). an even number as the sum :-

Favourable outcomes to an even number as the sum on dice= (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 outcomes to an even number as the sum on dice =19

We know that

Probability of an event = P(E)

= Number of favourable outcomes/Total number of possible outcomes

=> Probability of getting an an even number as the sum on the dice

= 19/36

= 19/36

Probability of getting an even number as the sum on the dice = 19/36

d). a multiple of 5 as the sum :-

Number of favourable outcomes to the multiple of 5 as the sum on dice = (1,4),,(2,3),(3,2),(4,1),(4,6),(5,5),(6,4)

Total Number of favourable outcomes to the multiple of 5 as the sum on dice =7

We know that

Probability of an event = P(E)

= Number of favourable outcomes/Total number of possible outcomes

=> Probability of getting multiple of 5 as the sum on dice = 7/36

e). a multiple of 7 as the sum :-

Favourable outcomes = (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)

Total number of favourable outcomes = 6

probability of getting a multiple of 7 as the sum =

=6/36

=1/6

Probability of getting a multiple of 7 as the sum on the dice = 1/6

f). a multiple of 3 as the sum :-

Favourable outcomes = (1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3),(6,6)

Total number of favourable outcomes = 12

Probability of getting a multiple of 3 as the sum =

=12/36

=1/3

Probability of getting a multiple of 3 as the sum on the dice = 1/3

g). a sum more than 7:-

Favourable outcomes = (2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)

Total number of favourable outcomes = 15

Probability of getting a number more than 7

=15/36

=5/12

Probability of getting a number more than 7 on the dice = 5/12

h). a sum greater than 9:-

Favourable outcomes = (4,6),(5,5),(5,6),(6,4),(6,5),(6,6)

Total number of favourable outcomes = 6

Probability of getting a number more than 9

=6/36

=1/6

Probability of getting a number more than 9 on the dice = 1/6

i). neither the sum 9 nor the sum 11:-

Favourable outcomes = (1,1),(1,2),(1,3),(1,4),(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),(4,1),(4,2),(4,3),(4,4),(4,6),(5,1),(5,2)(5,3),(5,5),(6,1),(6,2)(6,4),(6,6)

Total number of favourable outcomes 30

Probability of getting a number the sum 9 nor the sum 11

=30/36

=5/6

Probability of getting a number the sum 9 nor the sum 11 =5/6

j) a sum less than 6 :-

Probability of getting a number is less than 6 = 11/36

k). a sum less than 7:-

Probability of getting a sum less than 7

= 15/36

=5/12

l). a sum more than 7 :-

Probability of getting a sum more than 7

= 15/36

=5/12

m). a multiple of 3 on one dice:-

Probability of getting a number multiple of 3

= 20/36

= 5/9

n). a multiple of 2 on one dice:-

Probability of getting a multiple of 2 on one die

= 27/36

=3/4

o). a multiple of 5 on one dice :-

Probability of getting a number multiple of 5 on one die

= 11/36

p). a multiple of 2 on one dice and a multiple of 3 on the other :-

Probability of getting multiple of 2 on one dice and multiple of 3 on the other die

= 11/36

q). a doublet :-

= 6/36

= 1/6

r). a doublet of even. number:-

= 3/36

=1/12

s). a doublet of odd number:-

=3/36

=1/12

t). a doublet of prime .number :-

=3/36

=1/12

u). a number other than 5 on any dice :-

= 25/36

v). a number other than 3 on .any dice:-

25/36

w). the sum equal to 12:-

= 1/36

x). the sum more than equal to 10 :-

=6/36

=1/6

y). the sum less than or. equal to 10 :-

=33/36

=11/12

z).the sum as a prime number:-

=15/36

= 5/12

Answered by anita256874
2

A) 1/2

B) 1/2

C) 19/36

D) 7/36

E) 1/6

F) 1/3

G) 5/12

H) 1/6

I) 5/6

J) 11/36

K) 5/12

L) 5/12

M)5/9

N)3/4

O)11/36

P)11/36

Q)1/6

R)1/12

S)1/12

T)1/12

U)25/36

V)25/36

W)1/36

X)1/6

Y)11/12

Z)5/12

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