Here total no. of marbles is 20 only on first selection, but in the next selection we are left with only 19 marbles instead of 20 so denominator becomes 19. However if we need to take out one more than the denominator will become 18.

Since probability = \(\frac{favorable outcome}{total number of outcomes}\)

Hope it clears

We have 3 possible scenarios: 1) 2 reds, 2) 2 blues, and 3) 2 greens, thus:

Number of ways to select 2 reds is 7C2 = 7!/(2! x 5!) = (7 x 6)/2 = 21.

Number of ways to select 2 blues is 5C2 = 5!/(2! x 3!) = (5 x 4)/2! = 10.

Number of ways to select 2 greens is 8C2 = 8!/(2! x 6!) = (8 x 7)/2! = 28.

Number of ways to select any 2 marbles from 20 is 20C2 = 20!/(2! x 18!) = (20 x 19)/2 = 190.

Therefore, P(picking a pair of same-color marbles) = (21 + 10 + 28)/190 = 59/190.

Answer: D

Suppose the number of each marble was same. Would you have then taken into consideration the selection of first marble. I am asking because there was a similar question on gumdrops where , the number of different types of gumdrops were same and you had not taken ionto consideration , the probability of picking the first gumdrop and solved the question.

Yes..

say this question had 5 of each, then answer would have been .. 1*4/14
it would be same as (5/15)(4/14)+(5/15)(4/14)+(5/15)(4/14)= 1*4/14

P(matching marbles) = P(both are red OR both are blue OR both are green)
= P(red 1st and red 2nd OR blue 1st and blue 2nd OR green 1st and green 2nd)
= P(red 1st and red 2nd) + P(blue 1st and blue 2nd) + P(green 1st and green 2nd)
= [P(red 1st) x P(red 2nd)] + [P(blue 1st) x P(blue 2nd)] + [P(green 1st) x P(green 2nd)]
= [7/20 x 6/19] + [5/20 x 4/19] + [8/20 x 7/19]
= 42/380 + 20/380 + 56/380
= 118/380
= 59/190

Answer: D