Question

John had a bag of marbles as shown below.

Answer 1

Answer:

$\sf A) \quad \dfrac{9}{100}$

Step-by-step explanation:

Given:

• 3 green marbles
• 4 blue marbles
• 2 red marbles
• 1 yellow marble

Total number of marbles = 3 + 4 + 2 + 1 = 10

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

Probability of selecting a green marble on first pick:

$\implies \sf P(green\:marble)=\dfrac{3}{10}$

Probability of selecting a green marble on second pick:

$\implies \sf P(green\:marble)=\dfrac{3}{10}$

(as the 1st pick was replaced)

Therefore,

$\implies \textsf{P(green marble) and P(green marble)} \sf =\dfrac{3}{10} \times \dfrac{3}{10}=\dfrac{9}{100}$