Math / Data & Statistics

Question\begin{tabular}{|c|c|} \hline What are the chances of them adding up to 7 ? & 0.167 \\ \hline What are the chances of them adding up to 6 or $8 ?$ & 0.42 \\ \hline What are the chances of them adding up to a prime number? & 0.42 \\ \hline What are the chances of rolling a double (same number on each die)? & 0.167 \\ \hline What are the chances of rolling snake-eyes twice in a row? & 0.42 \\ \hline What are the chances of the dice adding up to an even number? & 0.5 \\ \hline What are the chances that both dice will be even numbers? & 0 \\ \hline \end{tabular}

Studdy Solution

STEP 1

What is this asking? We're checking which probabilities in this table about rolling two dice are correct! Watch out! Don't forget that each die has 6 sides, numbered 1 to 6!
Also, remember that probabilities must be between 0 and 1, inclusive.

STEP 2

1. Analyze the probability of the dice adding up to 7.
2. Analyze the probability of the dice adding up to 6 or 8.
3. Analyze the probability of the dice adding up to a prime number.
4. Analyze the probability of rolling a double.
5. Analyze the probability of rolling snake-eyes twice in a row.
6. Analyze the probability of the dice adding up to an even number.
7. Analyze the probability of both dice being even numbers.

STEP 3

There are $6 \cdot 6 = 36$ possible outcomes when rolling two dice.
Let's find the combinations that add up to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
There are 6 such combinations.

STEP 4

The probability of rolling a sum of 7 is the number of favorable outcomes divided by the total number of outcomes: $\frac{6}{36} = \frac{1}{6} \approx 0.167$ .
This matches the table!

STEP 5

Combinations that sum to 6: (1,5), (2,4), (3,3), (4,2), (5,1).
That's 5 combinations.

STEP 6

Combinations that sum to 8: (2,6), (3,5), (4,4), (5,3), (6,2).
That's 5 combinations.

STEP 7

There are $5 + 5 = \textbf{10}$ combinations that result in a sum of 6 or 8.
The probability is $\frac{10}{36} = \frac{5}{18} \approx 0.278$ .
The table says 0.42, so it's incorrect!

STEP 8

The possible sums are 2 through 12.
The prime numbers in this range are 2, 3, 5, 7, 11.

STEP 9

Let's count the combinations for each prime sum: 2:(1,1); 3:(1,2),(2,1); 5:(1,4),(2,3),(3,2),(4,1); 7:(1,6),(2,5),(3,4),(4,3),(5,2),(6,1); 11:(5,6),(6,5).
That's $1 + 2 + 4 + 6 + 2 = \textbf{15}$ combinations.

STEP 10

The probability is $\frac{15}{36} = \frac{5}{12} \approx 0.417$ .
Close enough to 0.42 in the table, so we'll call this correct!

STEP 11

Doubles are (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6).
There are 6 doubles.

STEP 12

The probability is $\frac{6}{36} = \frac{1}{6} \approx 0.167$ .
This matches the table!

STEP 13

Snake-eyes is (1,1).
The probability of rolling snake-eyes once is $\frac{1}{36}$ .

STEP 14

The probability of rolling snake-eyes twice in a row is $\frac{1}{36} \cdot \frac{1}{36} = \frac{1}{1296} \approx 0.00077$ .
The table's 0.42 is way off!

STEP 15

The possible even sums are 2, 4, 6, 8, 10, and 12.

STEP 16

We already know there are 1, 3, 5, 5, 3, and 1 combinations for each of these sums, respectively (from previous steps).
That's $1 + 3 + 5 + 5 + 3 + 1 = \textbf{18}$ combinations.

STEP 17

The probability is $\frac{18}{36} = \frac{1}{2} = 0.5$ .
This matches the table!

STEP 18

Even numbers on a die are 2, 4, and 6.
There are 3 even numbers on each die.

STEP 19

The combinations where both dice are even are (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), and (6,6).
That's $3 \cdot 3 = \textbf{9}$ combinations.

STEP 20

The probability is $\frac{9}{36} = \frac{1}{4} = 0.25$ .
The table says 0, so it's incorrect!

STEP 21

The incorrect probabilities in the table are: the chances of them adding up to 6 or 8 (should be approximately 0.278), the chances of rolling snake-eyes twice in a row (should be approximately 0.00077), and the chances that both dice will be even numbers (should be 0.25).

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Studdy Solution

STEP 1

What is this asking? We're checking which probabilities in this table about rolling two dice are correct! Watch out! Don't forget that each die has 6 sides, numbered 1 to 6!Also, remember that probabilities must be between 0 and 1, inclusive.

STEP 2

STEP 3

There are 6⋅6=366 \cdot 6 = 366⋅6=36 possible outcomes when rolling two dice.Let's **find** the combinations that add up to **7**: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).There are **6** such combinations.

STEP 4

The probability of rolling a sum of 7 is the number of favorable outcomes divided by the total number of outcomes: 636=16≈0.167\frac{6}{36} = \frac{1}{6} \approx 0.167366​=61​≈0.167.This matches the table!

STEP 5

**Combinations that sum to 6**: (1,5), (2,4), (3,3), (4,2), (5,1).That's **5** combinations.

STEP 6

**Combinations that sum to 8**: (2,6), (3,5), (4,4), (5,3), (6,2).That's **5** combinations.

STEP 7

There are 5+5=105 + 5 = \textbf{10}5+5=10 combinations that result in a sum of 6 or 8.The probability is 1036=518≈0.278\frac{10}{36} = \frac{5}{18} \approx 0.2783610​=185​≈0.278.The table says 0.42, so it's incorrect!

STEP 8

The possible sums are 2 through 12.The **prime numbers** in this range are 2, 3, 5, 7, 11.

STEP 9

Let's **count the combinations** for each prime sum: 2:(1,1); 3:(1,2),(2,1); 5:(1,4),(2,3),(3,2),(4,1); 7:(1,6),(2,5),(3,4),(4,3),(5,2),(6,1); 11:(5,6),(6,5).That's 1+2+4+6+2=151 + 2 + 4 + 6 + 2 = \textbf{15}1+2+4+6+2=15 combinations.

STEP 10

The probability is 1536=512≈0.417\frac{15}{36} = \frac{5}{12} \approx 0.4173615​=125​≈0.417.Close enough to 0.42 in the table, so we'll call this correct!

STEP 11

Doubles are (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6).There are **6** doubles.

STEP 12

The probability is 636=16≈0.167\frac{6}{36} = \frac{1}{6} \approx 0.167366​=61​≈0.167.This matches the table!

STEP 13

Snake-eyes is (1,1).The probability of rolling snake-eyes once is 136\frac{1}{36}361​.

STEP 14

The probability of rolling snake-eyes twice in a row is 136⋅136=11296≈0.00077\frac{1}{36} \cdot \frac{1}{36} = \frac{1}{1296} \approx 0.00077361​⋅361​=12961​≈0.00077.The table's 0.42 is way off!

STEP 15

The possible even sums are 2, 4, 6, 8, 10, and 12.

STEP 16

We already know there are 1, 3, 5, 5, 3, and 1 combinations for each of these sums, respectively (from previous steps).That's 1+3+5+5+3+1=181 + 3 + 5 + 5 + 3 + 1 = \textbf{18}1+3+5+5+3+1=18 combinations.

STEP 17

The probability is 1836=12=0.5\frac{18}{36} = \frac{1}{2} = 0.53618​=21​=0.5.This matches the table!

STEP 18

Even numbers on a die are 2, 4, and 6.There are 3 even numbers on each die.

STEP 19

The combinations where both dice are even are (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), and (6,6).That's 3⋅3=93 \cdot 3 = \textbf{9}3⋅3=9 combinations.

STEP 20

The probability is 936=14=0.25\frac{9}{36} = \frac{1}{4} = 0.25369​=41​=0.25.The table says 0, so it's incorrect!

STEP 21

The incorrect probabilities in the table are: the chances of them adding up to 6 or 8 (should be approximately 0.278), the chances of rolling snake-eyes twice in a row (should be approximately 0.00077), and the chances that both dice will be even numbers (should be 0.25).

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What is this asking? We're checking which probabilities in this table about rolling two dice are correct! Watch out! Don't forget that each die has 6 sides, numbered 1 to 6!
Also, remember that probabilities must be between 0 and 1, inclusive.

There are $6 \cdot 6 = 36$ possible outcomes when rolling two dice.
Let's find the combinations that add up to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
There are 6 such combinations.

The probability of rolling a sum of 7 is the number of favorable outcomes divided by the total number of outcomes: $\frac{6}{36} = \frac{1}{6} \approx 0.167$ .
This matches the table!

Combinations that sum to 6: (1,5), (2,4), (3,3), (4,2), (5,1).
That's 5 combinations.

Combinations that sum to 8: (2,6), (3,5), (4,4), (5,3), (6,2).
That's 5 combinations.

There are $5 + 5 = \textbf{10}$ combinations that result in a sum of 6 or 8.
The probability is $\frac{10}{36} = \frac{5}{18} \approx 0.278$ .
The table says 0.42, so it's incorrect!

The possible sums are 2 through 12.
The prime numbers in this range are 2, 3, 5, 7, 11.

Let's count the combinations for each prime sum: 2:(1,1); 3:(1,2),(2,1); 5:(1,4),(2,3),(3,2),(4,1); 7:(1,6),(2,5),(3,4),(4,3),(5,2),(6,1); 11:(5,6),(6,5).
That's $1 + 2 + 4 + 6 + 2 = \textbf{15}$ combinations.

The probability is $\frac{15}{36} = \frac{5}{12} \approx 0.417$ .
Close enough to 0.42 in the table, so we'll call this correct!

Doubles are (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6).
There are 6 doubles.

The probability is $\frac{6}{36} = \frac{1}{6} \approx 0.167$ .
This matches the table!

Snake-eyes is (1,1).
The probability of rolling snake-eyes once is $\frac{1}{36}$ .

The probability of rolling snake-eyes twice in a row is $\frac{1}{36} \cdot \frac{1}{36} = \frac{1}{1296} \approx 0.00077$ .
The table's 0.42 is way off!

We already know there are 1, 3, 5, 5, 3, and 1 combinations for each of these sums, respectively (from previous steps).
That's $1 + 3 + 5 + 5 + 3 + 1 = \textbf{18}$ combinations.

The probability is $\frac{18}{36} = \frac{1}{2} = 0.5$ .
This matches the table!

Even numbers on a die are 2, 4, and 6.
There are 3 even numbers on each die.

The combinations where both dice are even are (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), and (6,6).
That's $3 \cdot 3 = \textbf{9}$ combinations.

The probability is $\frac{9}{36} = \frac{1}{4} = 0.25$ .
The table says 0, so it's incorrect!