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John had a bag of marbles as shown below.
What is the probability that John selects a green marble, replaces it, and then selects a
green marble?
3 green marbles
4 blue marbles
2 red marbles
1 yellow marble

A.9/100
B.1/ 10
C.1/15
D.1/50

Respuesta :

semsee45

Answer:

[tex]\sf A) \quad \dfrac{9}{100}[/tex]

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

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

Probability of selecting a green marble on first pick:

[tex]\implies \sf P(green\:marble)=\dfrac{3}{10}[/tex]

Probability of selecting a green marble on second pick:

[tex]\implies \sf P(green\:marble)=\dfrac{3}{10}[/tex]

(as the 1st pick was replaced)

Therefore,

[tex]\implies \textsf{P(green marble) and P(green marble)} \sf =\dfrac{3}{10} \times \dfrac{3}{10}=\dfrac{9}{100}[/tex]

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