Question

A jar contains 30 marbles. It has 10 red, 6 black and 14 green marbles. Two marbles are drawn, the first is not returned before the second one is drawn. What is the probability that both marbles are green?
P(Both Green) =
14 / 169


P(Both Green) =
91 / 435


P(Both Green) =
7 / 15


P(Both Green) =
49 / 225

Answer 1

Answer:

The correct answer option is P (both green) = 91 / 435

Step-by-step explanation:

We are given that in a jar containing 30 marbles, 10 are red, 6 are black and 14 are green.

Two marbles are drawn and the second one is drawn without returning the first marble and we are to find the probability of getting green marbles both time.

P (both green) = $\frac { 1 4 } { 3 0 } \times \frac { 1 3 } { 2 9 }$ = 91 / 435