A bag contains 10 red marbles and 5 blue marbles. You randomly select 3 marbles from the bag. What is the probability that all 3 marbles are blue when (a) you replace each marble before selecting the next one and (b) you do not replace each marble before selecting the next one?

(a) 1/3 (b) 1/3
(a) 1/27 (b) 1/3
(a) 1/27 (b) 2191
(a) 1/3 (b) 2/91

A.) With replacement = 1/27

B.) Without replacement = 2/91

Explanation:

Given the following :

Red marbles = 10

Blue marbles = 5

Probability that all 3 marbles are blue;

A.) With replacement :

Total no marbles = 10 + 5 = 15

P(First pick equals blue) = 5/15 = 1/3

P( second pick equals blue) = 5/15 = 1/3

P( third pick equals blue) = 5/15 = 1/3

P(blue blue blue) = (1/3) × (1/3) × (1/3) = 1/9

B.) Without replacement :

P(First pick equals blue) = 5/15 = 1/3

P( second pick equals blue) = 4/14

P( third pick equals blue) = 3/13

P(blue blue blue) = (1/3) × (4/14) × (3/13 )

= 12/546 = 2/91

answer; i believe that the correct answer is ( troxel v. granville);

the median of the data is 4.

step-by-step explanation:

the median means the middle number of a given number.

to decide which one is the median, arrange in ascending or descending order. but sometimes, you may find two(2) number in the middle, but it's easy, just add them together and divide by 2.

so,   determine the median of the data. 2, 3, 3, 4, 5, 7, 9

it is already arranged, count 3 times forward and 3 backward and the one left is your answer.

2,3,3,4,5,7,9

our answer is 4.

hope this !