5.3 Multiplication Rule and Conditional Probability

If the probability of event A is 0.4 and the probability of event B is 0.3, what is the probability of both events occurring if both events are independent?

• 0.12

• 0.5

• 0.25

• 0.07

Answer explanation

The probability of both independent events occurring is the product of their individual probabilities, so 0.4 * 0.3 = 0.12.

Tags

CCSS.HSS.CP.A.2

CCSS.HSS.CP.A.4

If the probability of rain on any given day is 0.2, what is the probability of no rain for 3 consecutive days?

• 0.6

• 0.3

• 0.512

• 0.8

Answer explanation

The probability of no rain for 3 consecutive days is calculated as (1-0.2)^3 = 0.512, hence the correct answer is 0.512.

If P(A)=0.6 and the P(B|A)=0.8, what is P(A∩B)?

• 0.2

• 0.25

• 0.75

• 0.48

Answer explanation

To find P(A∩B), use the formula P(A∩B) = P(A) * P(B|A) = 0.6 * 0.8 = 0.48. Therefore, the correct answer is 0.48.

A box contains 4 blue marbles, 3 red marbles, and 2 green marbles. What is the probability of drawing a blue marble followed by a red marble without replacement?

• 2/7

• 3/10

• 1/3

• 1/6

Answer explanation

The probability of drawing a blue marble first is 4/9. After drawing a blue marble, there are 3 red marbles left out of 8 total marbles, giving a probability of 3/8. Multiplying these probabilities gives 1/6.

If P(A)=0.7 and the P(B)=0.5, what is P(A∪B) if A and B are independent events?

• 0.9

• 0.2

• 0.3

• 0.85

Answer explanation

When A and B are independent, P(A∪B) = P(A) + P(B) - P(A) * P(B) = 0.7 + 0.5 - (0.7 * 0.5) = 0.85

Tags

CCSS.HSS.CP.B.7

A jar contains 10 red candies and 5 green candies. What is the probability of drawing a green candy followed by a red candy without replacement?

• 3/10

• 5/21

• 1/5

• 2/15

Answer explanation

The probability of drawing a green candy first is 5/15. After removing the green candy, there are 10 red candy out of 14 candy, then drawing a red candy is 10/14. Multiplying these probabilities gives 5/21.

A box contains 6 white balls and 4 black balls. What is the probability of drawing a white ball followed by a black ball without replacement?

• 2/5

• 4/15

• 5/15

• 3/10

Answer explanation

The probability of drawing a white ball first is 6/10. After removing one white ball, the probability of drawing a black ball is 4/9. Multiplying these probabilities gives 4/15.

If the probability of event A is 0.3 and the probability of event B given A has occurred is 0.6, what is the conditional probability of event A given B?

• 0.5

• 0.4

• 0.3

• 0.7

Answer explanation

The conditional probability of event A given B is calculated using Bayes' theorem:

P(A|B) =[ P(B|A) * P(A)] / P(B)

Substituting the given values, we get P(A|B) =( 0.6 * 0.3) / P(B)

Since P(B) = P(B|A) * P(A) + P(B|A') * P(A') =(0.6 * 0.3) + (0.6 * 0.7) = 0.6,

then P(A|B)=(0.6 * 0.3) / 0.6 = 0.3.

Tags

CCSS.HSS.CP.A.3

CCSS.HSS.CP.A.5

Given that A and B are independent event with the probability of event A is 0.8 and the probability of event B is 0.4, what is the probability of both events not occurring?

• 0.32

• 0.2

• 0.12

• 0.6

Answer explanation

If A and B are independent, then P(A n B)=P(A)*P(B).

The probability of both events not occurring is P(A' n B'), so

P(A' n B')=P(A')*P(B')=(1-0.8)*(1-0.4)=0.12

Tags

CCSS.HSS.CP.B.7

A box contains 3 yellow balls, 2 green balls, and 1 red ball. Two balls are taken without replacement. What is the probability of drawing a yellow ball given that the first ball is a green ball?

• 1/5

• 1/6

• 2/5

• 1/3

Answer explanation

P(Yellow|Green)=P(Green n Yellow)/P(Green)

P(Green n Yellow)=2/6*3/5=1/5

P(Green)=2/6=1/3

Then P(Yellow|Green)=(1/5)/(1/3)=3/5

Tags

CCSS.HSS.CP.B.6