A game has 2 spinners. Spinner #1 has a probability of landing red of $2 / 3$ . And, spinner#2 has a probability of landing red of $1 / 5$ .
What is the probability spinner#1 lands red AND spinner#2 does NOT land red?

2 / 15

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8 / 15

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13 / 15

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1 / 5

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3 / 5

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Solution

The correct option is B $8 / 15$
probability of spinner#1 lands red AND spinner#2 doesnot land red is $2 / 3 \times (1 - 1 / 5) = 2 / 3 \times 4 / 5 = 8 / 15$

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A game has 2 spinners. Spinner #1 has a probability of landing red of 2/3. And, spinner#2 has a probability of landing red of 1/5.What is the probability spinner#1 lands red AND spinner#2 does NOT land red?

The correct option is B 8/15probability of spinner#1 lands red AND spinner#2 doesnot land red is 2/3×(1−1/5)=2/3×4/5=8/15

A game has 2 spinners. Spinner #1 has a probability of landing red of $2 / 3$ . And, spinner#2 has a probability of landing red of $1 / 5$ .
What is the probability spinner#1 lands red AND spinner#2 does NOT land red?

The correct option is B $8 / 15$
probability of spinner#1 lands red AND spinner#2 doesnot land red is $2 / 3 \times (1 - 1 / 5) = 2 / 3 \times 4 / 5 = 8 / 15$