Question

A bag contains 10 blue, 6 green, and 4 red marbles. You choose a marble with one hand and then a second marble with the other hand. What is the probability that you will first choose a red marble followed by a green marble?

Answer 1

Answer:

https://brainly.com/question/3050696

check here 4 the answer, there is slight change in the wordings of the question but the answer is same....

Step-by-step explanation:

Answer 2

The probability that you will first choose a red marble followed by a green marble will be  $\dfrac{6}{19}$.

Explanation:

Given : A bag contains 10 blue, 6 green, and 4 red marbles.

Total marbles = 20

The probability that the first marble is red marble = $\dfrac{\text{No. of red marbles}}{\text{Total marbles}}$

$=\dfrac{4}{20}=\dfrac{1}{5}$

After drawing the first red marble , the total marbles left in the bag= 20-1=19

The total green marbles remain the same.

Then, the probability that you will first choose a red marble followed by a green marble will be :

$\dfrac{\text{No. of green marbles left}}{\text{Total remaining marnles}}$

$=\dfrac{6}{19}$

Hence, the probability that you will first choose a red marble followed by a green marble will be  $\dfrac{6}{19}$.

# Learn more :

If the probability of a win is 0.12 and the probability of a draw is 0.6, what is the probability of a loss?

https://brainly.in/question/7962977