Abag contains only red and blue marbles. yasmine takes one marble at random from the bag. the probability that she takes a red marble is 1 in 5. yasmine returns the marble to the bag and adds five more red marbles to the bag. the probability that she takes one red marble at random is now 1 in 3. how many red marbles were originally in the bag?

Initially there were 5 red marbles.

Explanation:

Let the original number of red marbles = x

Let the original number of blue marbles = y

Then \[x/x+y = 1/5\]

=> \[5x = x + y\]

=> \[4x = y\]

After adding five more red marbles, (x+5)/(x+y+5) = 1/3

=> \[3x + 15 = x+ y + 5\]

=> \[2x + 10 = y\]

Solving, \[4x = 2x + 10\]

=> \[4x -2x = 10\]

Or x = 5

y = 4x = 20

So initially there were 5 red marbles.

Verifying the result,

Initial condition: \[5/(20+5) = 1/5\]

Final condition: \[(5+5)/(20+5+5) = 1/3\]

imma come back to you?

explanation:

d. long term memory