The number of marbles of different colors stored in a hat is listed below:

8 red marbles
10 green marbles
6 blue marbles
without looking in the hat, tessa takes out a marble at random. she replaces the marble and then takes out another marble from the hat. what is the probability that tessa takes out a blue marble in both draws?

fraction 1 over 16

fraction 1 over 12

fraction 1 over 4

fraction 1 over 2

it's not 1/4 because i tried that.

Step-by-step explanation:

Given : The number of marbles of different colors stored in a hat is listed below:

8 red marbles

10 green marbles

6 blue marbles

To Find: What is the probability that Tessa takes out a blue marble in both draws?

Solution:

8 red marbles

10 green marbles

6 blue marbles

Total Marbles = 8+10+6= 24

Probability of getting blue marble on first draw =

Probability of getting blue marble on first draw =

Now the marble is replaced

Probability of getting blue marble on second draw =

Probability of getting blue marble on second draw =

So, probability of getting blue marble on both the draws =  

                                                                                               =  

                                                                                               =  

Hence the probability that Tessa takes out a blue marble in both draws is

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18