A bag contains 2 red,5 blue,6 yellow, and 3 green crayons. What is the probability of selecting a red crayon, then without replacing it selecting an odd number that is (red, then green)
A.0
B.0.25
C.01
D.0.025

Answer: D.0.025

Step-by-step explanation:

Given: A bag contains 2 red,5 blue,6 yellow, and 3 green crayons.

Total crayons: 2+5+6+3=16

Probability of getting red crayon= [tex]\dfrac{\text{Number of red crayons}}{\text{Total}}[/tex]

[tex]=\dfrac{2}{16}=\dfrac18[/tex]

Without replacement, the probability of getting green crayon = [tex]\dfrac{\text{Number of green crayons}}{\text{Total}-1}[/tex]

[tex]=\dfrac{3}{15}=\dfrac15[/tex]

Required probability P(red then green)= [tex]\dfrac18\times\dfrac15=\dfrac1{40}=0.025[/tex]

Hence, correct option is D.0.025

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