# Lesson 4 & 5 Quiz - Possibility Diagrams & Tree Diagrams

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Lesson 4 & 5 Quiz - Possibility Diagrams & Tree Diagrams. Please go through the lesson before attempting the quiz. This quiz has 15 questions. There will not be any time limit for this quiz and answer all questions. You need to answer at least 10 questions correctly in order to pass the quiz. Please acknowledge by entering you name before attempting the quiz. Good luck!

• 1.

### X = {3, 4, 5, 6} and Y = {7, 8, 9, 10}. An element x is selected randomly from X and an element y is selected from Y. The value of x + y is then recorded.What is the probability that x + y is odd?

• A.

1 / 8

• B.

1 / 6

• C.

1 / 4

• D.

1 / 2

D. 1 / 2
Explanation
Let E be the event that x + y is odd.

Using the possibility diagram, we can easily see that n(S) = 16 and n(E) = 8.

Therefore, P(E) = 8 / 16 = 1 / 2

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• 2.

### X = {3, 4, 5, 6} and Y = {7, 8, 9, 10}. An element x is selected randomly from X and an element y is selected from Y. The value of x + y is then recorded.What is the probability that x + y is even?

• A.

1 / 8

• B.

1 / 6

• C.

1 / 4

• D.

1 / 2

D. 1 / 2
Explanation
Let F be the event that x + y is even.

Using the possibility diagram, we can easily see that n(S) = 16 and n(F) = 8.

Therefore, P(F) = 8 / 16 = 1 / 2

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• 3.

### X = {3, 4, 5, 6} and Y = {7, 8, 9, 10}. An element x is selected randomly from X and an element y is selected from Y. The value of xy is then recorded.What is the probability that xy is a multiple of 3?

• A.

3 / 8

• B.

5 / 8

• C.

3 / 16

• D.

5 / 16

B. 5 / 8
Explanation
Let H be the event that xy is a multiple of 3.

Using the possibility diagram, we can easily see that n(S) = 16 and n(H) = 10.

Therefore, P(H) = 10 / 16 = 5 / 8

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• 4.

### X = {3, 4, 5, 6} and Y = {7, 8, 9, 10}. An element x is selected randomly from X and an element y is selected from Y. The value of xy is then recorded.What is the probability that xy is less than or equal to 30?

• A.

3 / 8

• B.

5 / 8

• C.

3 / 16

• D.

5 / 16

D. 5 / 16
Explanation
Let H be the event that xy is less than or equal to 30 .

Using the possibility diagram, we can easily see that n(S) = 16 and n(I) = 5.

Therefore, P(I) = 5 / 16 = 5 / 16

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• 5.

### In a game, the player throws a coin and a regular die simultaneously. If the coin shows a head, the player's score is the score on the die. If the coin shows a tail, then the player's score is twice the score on the die. What is the probability that the score is even?

• A.

1 / 4

• B.

1 / 2

• C.

3 / 4

• D.

1

C. 3 / 4
Explanation
Let E be the score is even.

Using the possibility diagram, we can easily see that n(S) = 12 and n(E) = 9.

Therefore, P(E) = 9 / 12 = 3 / 4

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• 6.

### In a game, the player throws a coin and a regular die simultaneously. If the coin shows a head, the player's score is the score on the die. If the coin shows a tail, then the player's score is twice the score on the die. What is the probability that the score is less than 6?

• A.

5 / 6

• B.

5 / 12

• C.

7 / 12

• D.

3 / 4

B. 5 / 12
Explanation
Let F be the score is even.

Using the possibility diagram, we can easily see that n(S) = 12 and n(F) = 5.

Therefore, P(F) = 5 / 12

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• 7.

### A coin is flipped 4 times. What is the value of n(S)?

• A.

4

• B.

8

• C.

12

• D.

16

D. 16
Explanation
Using a tree diagram, we can see that n(S) = 16.

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• 8.

### A coin is flipped 4 times. What is the probability of getting more than 2 heads?

• A.

3 / 8

• B.

5 / 8

• C.

3 / 16

• D.

5 / 16

D. 5 / 16
Explanation
Let F be the event of getting more than 2 heads.

Using a tree diagram, we can see that n(F) = 5.

Therefore, P(F) = 5 / 16

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• 9.

### A coin is flipped 4 times. What is the probability of getting all 4 coins land on the same side?

• A.

1 / 16

• B.

1 / 8

• C.

1 / 4

• D.

1 / 2

B. 1 / 8
Explanation
Let G be the event of all 4 coins land on the same side.

Using a tree diagram, we can see that n(G) = 2.
Note: 2 possible outcomes - All heads and all tails.

Therefore, P(G) = 2 / 16 = 1 / 8

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• 10.

### In a game, a player will have to throw a tetrahedral die and a coin at the same time. If he get a tail on the coin, the game will stop and he will receive the point based on the side facing down on the tetrahedral die. If he get a head on the coin, he will then throw the die again and he will receive the point based on the product of the 2 throws on the die. What are the number of different possible score a player can get?

• A.

4

• B.

8

• C.

9

• D.

16

C. 9
Explanation
From the tree diagram, we can see that the possible scores that the player can score = {1, 2, 3, 4, 6, 8, 9, 12, 16}

Therefore, number of possible scores = 9.

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• 11.

### In a game, a player will have to throw a tetrahedral die and a coin at the same time. If he get a tail on the coin, the game will stop and he will receive the point based on the side facing down on the tetrahedral die. If he get a head on the coin, he will then throw the die again and he will receive the point based on the product of the 2 throws on the die. What is the value of n(S)?

• A.

8

• B.

9

• C.

16

• D.

20

D. 20
Explanation
From the tree diagram, we can see that n(S) = 20.

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• 12.

### In a game, a player will have to throw a tetrahedral die and a coin at the same time. If he get a tail on the coin, the game will stop and he will receive the point based on the side facing down on the tetrahedral die. If he get a head on the coin, he will then throw the die again and he will receive the point based on the product of the 2 throws on the die. What is the probability of getting a score greater than 7?

• A.

3 / 8

• B.

3 / 10

• C.

3 / 16

• D.

3 / 20

B. 3 / 10
Explanation
Let E be the event of getting a score greater than 7.

From the tree diagram, we can see that n(S) = 20 and n(E) = 6.

Therefore, P(E) = 6 / 20 = 3 / 10

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• 13.

### In a game, a player will have to throw a tetrahedral die and a coin at the same time. If he get a tail on the coin, the game will stop and he will receive the point based on the side facing down on the tetrahedral die. If he get a head on the coin, he will then throw the die again and he will receive the point based on the product of the 2 throws on the die. If the player can only wins if he/she get above a certain value. What is the value if the game only allows 2 / 5 of the players to win?

• A.

4

• B.

5

• C.

6

• D.

7

B. 5
Explanation
Let F be the event that the player score more than 5 points.

From the tree diagram, we can see that n(S) = 20 and n(F)= 8.

Therefore, P(F) = 8 / 20 = 4 / 5

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• Current Version
• Mar 21, 2023
Quiz Edited by
ProProfs Editorial Team
• Mar 17, 2009
Quiz Created by
Enxian87

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