Question

A bag of marbles contains 7 red, 5 blue, 4 green, and 2 yellow marbles. Jon selects a marble, replaces it, then selects another marble. What is the probability that Jon selects a red marble and then a yellow marble?

Answer 1

Answer: about 0.0432 or 4.32%

Step-by-step explanation:

Given : A bag of marbles contains 7 red, 5 blue, 4 green, and 2 yellow marbles.

Total marbles = 7+5+4+2=18

Let R : Event of getting first marble as red .

Y= Event of getting second marble as yellow.

Jon selects a marble, replaces it, then selects another marble.

⇒Both events are independent .

Probability of getting first marble as red = $P(R)=\dfrac{\text{Number of red marbles}}{\text{Total marbles}}$

$\\\\=\dfrac{7}{18}$

Probability of getting second marble as yellow = $P(Y)=\dfrac{\text{Number of yellow marbles}}{\text{Total marbles}}$

$\\\\=\dfrac{2}{18}$

Now, the probability that Jon selects a red marble and then a yellow marble :

$P(R)\times P(Y)=\dfrac{7}{18}\times\dfrac{2}{18}\approx0.0432=4.32\%$  [ ∵ Event R and Y are independent .]

Hence, the probability that Jon selects a red marble and then a yellow marble is about 0.0432 or 4.32%.

Answer 2

For the red 7 in 20 for the yellow 2 in 20