Independent & Dependent Probability

Complete the tree diagram.

What's the answer for #1?

(a)  

Answer explanation

For each branch, the sum of all of the probabilities equals 1.

Tags

CCSS.7.SP.C.8B

Complete the tree diagram.

What's the answer for #2?

(a)  

Answer explanation

For each branch, the sum of all of the probabilities equals 1.

Tags

CCSS.7.SP.C.8B

Complete the tree diagram.

What's the answer for #3?

(a)  

Answer explanation

For each branch, the sum of all of the probabilities equals 1.

Tags

CCSS.7.SP.C.8B

Complete the tree diagram.

What's the answer for #4?

(a)  

Answer explanation

For each branch, the sum of all of the probabilities equals 1.

Tags

CCSS.7.SP.C.8B

Find the probability that a member of the armed forces is an officer.

Round 3 sig fig

(a)  

Answer explanation

This person can be a

navy officer (0.22 * 0.19) = 0.0418

army officer (0.47 * 0.15) = 0.0705

air force officer (0.31 * 0.21) = 0.0651

Since there are multiple occurrences, you add the probabilities.

0.1774 rounds to 0.177 (3 sig fig)

Find the probability that a member of the armed forces is not an officer in the navy.

Round 3 sig fig

(a)  

Answer explanation

P(officer in navy) = (0.22 * 0.19) = 0.0418

Think about the complement

1 - P(officer in navy)

1-0.0418 = 0.9582

0.958 (3 sig fig)

Tags

CCSS.HSS.CP.A.1

Find the probability that a member of the armed forces is not an army or air force officer.

(a)  

Answer explanation

P(army or air force officer)

(0.47 * 0.15) + (0.31 *0.21)

0.1356

Think complement

1-P(army or air force officer)

1 - 0.1356 = 0.8644

0.864 (3 sig fig)

Tags

CCSS.HSS.CP.A.1

You spin the spinner and flip a coin.

Find the probability of spinning an even number and flipping tails.

Type your answer as a decimal.

(a)  

Answer explanation

5/10 * 1/2 = 5/20 = 0.25

Tags

CCSS.HSS.CP.A.1

CCSS.HSS.CP.A.2

CCSS.HSS.CP.B.8

You spin the spinner and flip a coin.

Find the probability of spinning a white and not flipping tails.

Type your answer as a decimal.

(a)  

Answer explanation

5/10 * 1/2 = 5/20 = 0.25

Tags

CCSS.HSS.CP.B.7

A school has two photocopiers. On any given day, machine A has an 8% chance of malfunctioning and machine B has a 12% chance of malfunctioning. Determine the probability on any given day both machines will malfunction.

(a)  

Answer explanation

p(A and B malfunctioning)

0.08 * 0.12

0.0096

Tags

CCSS.HSS.CP.A.2

CCSS.HSS.CP.B.8