Question

A fair spinner has 12 equal sections: 3 red, 5 blue and 4 green.
It is spun twice.
What is the probability of getting not green then green?

Answer 1

Question:

A fair spinner has 12 equal sections: 3 red, 5 blue and 4 green.It is spun twice.What is the probability of getting not green then green?

Answer:

Probability of not getting green = $\dfrac{3}{12}$ = $\dfrac{1}{4}$

Probability of getting green = $\dfrac{9}{12}$ = $\dfrac{3}{4}$

$\dfrac{1}{4}$×$\dfrac{3}{4}$ = $\dfrac{3}{16}$

Answer 2

Answer:

27/144

Step-by-step explanation:

P(R,G) = 4/12 X 3/12 = 12/144

P(B,G) = 5/12 X 3/12 = 15/144

12+15 = 27

= 27/144