These marbles are placed in a bag and two
of them are randomly drawn.
GO
What is the probability of drawing two blue
marbles, if the first marble is NOT placed
back in the bag before the second draw?
Give your answer as a rational number, reduced to
simplest terms.
N
[ ?
Hint: Multiply the probability of the 1st Event by the
probability of the 2nd Event to get your answer.

2/9

Step-by-step explanation:

For the first draw:

We have,

Number of blue marbles (B)= 5

Total number of marbles (N)= 10

Probability of drawing a blue marble in first

draw [P(B)] = B/N

=5/10

=1/2

For the second draw when the initially drawn blue marble is not placed back in the bag:

We have,

Number of blue marbles (b) = 4

Total number of marbles (n) = 9

Probability of drawing a blue marble in the second draw [P(b)] = b/n

= 4/9

Finally,

Probability of drawing two blue marbles

[P(Bb)] = [P(B)] * [P(b)]

= 1/2 * 4/9

=2/9

Therefore, [P(Bb)] = 2/9

f(x)=2/3x+4/3 is the answer because is whose power in the senate is equal to the power of the speaker of the house?

step-by-step explanation:

.95

step-by-step explanation: