A standard fair dice is rolled twice. What is the probability of getting an odd number on the first roll and any number except 4 on the other?

Let,O and P be the events of getting odd number and any other number except 4 respectively.

n(S)=6

O=(1,3,5) so, n(O)=3

P=(1,2,3,5,6) so, n(P)=5

So,probability of getting an odd number and any other number except 4 will be

n(O)/n(S) × n(P)/n(S)

=3/6 × 5/6

=5/12

May be it's the answer I guess. Hope you're clear with it:-)

4+4+2x

or

2x+8

step-by-step explanation:

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -