Question

PLZ HURRY IT'S URGENT!!!

You randomly draw marbles from a bag containing 4 blue and 2 green marbles, without replacing the marbles between draws. What is the probability of drawing a green marble on both your first and second draws?
A. 1/2⋅1/4
B. 2/6⋅1/5
C. 2/6⋅1/6
D. 2/6⋅2/6

Answer 1

B)
`(2)/(6)* (1)/(5)`

Explanation:

Number of blue marbles =
`N(blue)=4`

Number of green marbles=
`N(green)=2`

Total number of balls=
`N=N(blue)+N(green)=4+2=6`

Probability of drawing a green marble on first draw=
`P(green_1)=(N(green))/(N)=(2)/(6)`

After drawing a green marble a 2nd draw is made without replacement.

So, now
`N(green)=2-1=1` and
`N=6-1=5`

Probability of drawing a green marble on 2nd draw=
`(N(green))/(N)=P(green_2)=(1)/(5)`

∴ Probability of drawing a green marble on both your first and second draws=
`P(green_1)* P(green_2)=(2)/(6)* (1)/(5)`