Question

There are 10 blue marbles, 4 black marbles, 5 white marbles, and 6 red marbles in a box. If two marbles are drawn at random without replacement, what is the probability that both marbles removed are not blue?​

Answer 1

Answer:

7/20

Step-by-step explanation:

10(blue) + 4(black) + 5(white) + 6(red) = 25(total)

25(total)-10(blue)=15(not blue)

not blue/all = 15/25 = 3/5 = probability that first marble removed are not blue

If we remove 1 marble;

25-1 = 24 (total) , 15-1=14(not blue)

second not blue/all = 14/24 = 7/12 = probability that second marble removed are not blue

probability that both marbles removed are not blue : 3/5. 7/12 = 21/60

21/60 = 7/20

Hope this helps ^-^

Answer 2

Answer:

7/20

Step-by-step explanation:

10 blue marbles, 4 black marbles, 5 white marbles, and 6 red marbles = 25 marbles

25-10 = 15

There are 15 not blue marbles

P( not blue ) = not blue marbles / total = 15/25 = 3/5

keep the not blue marble

25-1 = 24 marbles left

15-1 = 14 not blue marbles

P( not blue ) = not blue marbles / total = 14/24 = 7/12

P( not blue, keep, not blue) = 3/5 * 7/12 = 21/60 = 7/20