Mathematics: Plz hurry it s ! 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

6 in sqStep-by-step explanation:...

Plz hurry it s ! you randomly draw marbles from a

aylineorozco836

10.01.2022

B) $\frac{2}{6}\times \frac{1}{5}$

Step-by-step 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= $P(green_1)=\frac{N(green)}{N}=\frac{2}{6}$

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= $\frac{N(green)}{N}=P(green_2)=\frac{1}{5}$

∴ Probability of drawing a green marble on both your first and second draws= $P(green_1)\times P(green_2)=\frac{2}{6}\times \frac{1}{5}$

0,0(0 оценок)