Abag contains 10 green marbles and 4 yellow marbles. if two marbles are chosen at random, one at a time and without replacement, what is the probability of getting two green marbles?

 The required probability of getting two green marbles is 49.45%.

Step-by-step explanation:  Given that a bag contains 10 green marbles and 4 yellow marbles. Two marbles are chosen at random, one at a time and without replacement.

We are to find the probability of getting two green marbles.

Let S denote the sample space for the experiment of choosing a marble from the bag and A denote the event of getting a green marble.

The, n(S) = 10 + 4 = 14   and   n(A) = 10.

So, the probability of event A will be

After getting one green marble and not replacing, let S' denote the sample space for the experiment of choosing a marble from the bag

and

let B denote the event of getting another green marble.

Then, n(S') = 14 - 1 = 13    and    n(B) = 10 - 1 = 9.

Then, the probability of getting two green marbles is given by

Thus, the required probability of getting two marbles is 49.45%.

i would say c is the best answer, hope it's right!

good luck! =)

cutie~

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.