A bag contains 10 balls: 4 white and 6 black.
A ball is removed at random and replaced; a second ball is then removed.
What is the probability that the balls removed were the same colour?
(Answer is in its simplest form.)

first case

Step-by-step explanation:

now for ,

the probability that the balls removed were the same colour =P(w)×P'(w)+P'(b)×P(b)

=7/15

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752