Math / Data & Statistics

Question2. Miguel is comparing the probabilities of two spinners. Select all the true statements that Miguel made. (A) Spinner $A$ is twice as likely to land on 1 as spinner $B$ is. (B) The probability that spinner $B$ lands on an odd number is $\frac{1}{2}$ . (c) The probability that spinner $A$ lands on 2 is $\frac{1}{4}$ . (1)
Spinner $A$ is not a fair spinner. Spinner $A$ and spinner $B$ are equally likely to land on an odd number.

Studdy Solution

STEP 1

1. Spinner $A$ has six equally likely sections: $2, 1, 1, 3, 2, 3$ .
2. Spinner $B$ has four equally likely sections: $4, 1, 3, 2$ .
3. Probabilities are calculated based on the number of favorable outcomes over the total number of outcomes.

STEP 2

1. Calculate the probability of landing on specific numbers for both spinners.
2. Evaluate each statement based on the calculated probabilities.

STEP 3

Calculate the probability that spinner $A$ lands on 1.
Spinner $A$ has two sections labeled 1 out of six total sections.
$P(A \text{ lands on } 1) = \frac{2}{6} = \frac{1}{3}$

STEP 4

Calculate the probability that spinner $B$ lands on 1.
Spinner $B$ has one section labeled 1 out of four total sections.
$P(B \text{ lands on } 1) = \frac{1}{4}$

STEP 5

Compare the probabilities for statement (A).
$\frac{1}{3} \text{ is not twice } \frac{1}{4}$
Statement (A) is false.

STEP 6

Calculate the probability that spinner $B$ lands on an odd number.
Odd numbers on spinner $B$ are 1 and 3. There are two odd sections out of four.
$P(B \text{ lands on odd}) = \frac{2}{4} = \frac{1}{2}$
Statement (B) is true.

STEP 7

Calculate the probability that spinner $A$ lands on 2.
Spinner $A$ has two sections labeled 2 out of six total sections.
$P(A \text{ lands on } 2) = \frac{2}{6} = \frac{1}{3}$
Statement (C) is false because $\frac{1}{3} \neq \frac{1}{4}$ .

STEP 8

Determine if spinner $A$ is fair.
A fair spinner would have equal probabilities for each section, which is not the case here.
Spinner $A$ is not fair.

STEP 9

Compare the probability of landing on an odd number for both spinners.
Spinner $A$ odd numbers: 1, 1, 3, 3 (4 out of 6 sections).
$P(A \text{ lands on odd}) = \frac{4}{6} = \frac{2}{3}$
Spinner $B$ odd numbers: 1, 3 (2 out of 4 sections).
$P(B \text{ lands on odd}) = \frac{2}{4} = \frac{1}{2}$
Spinner $A$ and spinner $B$ are not equally likely to land on an odd number.
The true statements are: - (B) The probability that spinner $B$ lands on an odd number is $\frac{1}{2}$ . - Spinner $A$ is not a fair spinner.

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Studdy Solution

STEP 1

1. Spinner $A$ has six equally likely sections: $2, 1, 1, 3, 2, 3$ .
2. Spinner $B$ has four equally likely sections: $4, 1, 3, 2$ .
3. Probabilities are calculated based on the number of favorable outcomes over the total number of outcomes.

STEP 2

1. Calculate the probability of landing on specific numbers for both spinners.
2. Evaluate each statement based on the calculated probabilities.

STEP 3

Calculate the probability that spinner $A$ lands on 1.
Spinner $A$ has two sections labeled 1 out of six total sections.
$P(A \text{ lands on } 1) = \frac{2}{6} = \frac{1}{3}$

STEP 4

Calculate the probability that spinner $B$ lands on 1.
Spinner $B$ has one section labeled 1 out of four total sections.
$P(B \text{ lands on } 1) = \frac{1}{4}$

STEP 5

Compare the probabilities for statement (A).
$\frac{1}{3} \text{ is not twice } \frac{1}{4}$
Statement (A) is false.

STEP 6

Calculate the probability that spinner $B$ lands on an odd number.
Odd numbers on spinner $B$ are 1 and 3. There are two odd sections out of four.
$P(B \text{ lands on odd}) = \frac{2}{4} = \frac{1}{2}$
Statement (B) is true.

STEP 7

Calculate the probability that spinner $A$ lands on 2.
Spinner $A$ has two sections labeled 2 out of six total sections.
$P(A \text{ lands on } 2) = \frac{2}{6} = \frac{1}{3}$
Statement (C) is false because $\frac{1}{3} \neq \frac{1}{4}$ .

STEP 8

Determine if spinner $A$ is fair.
A fair spinner would have equal probabilities for each section, which is not the case here.
Spinner $A$ is not fair.

STEP 9

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Studdy Solution

STEP 1

1. Spinner A A A has six equally likely sections: 2,1,1,3,2,3 2, 1, 1, 3, 2, 3 2,1,1,3,2,3.2. Spinner B B B has four equally likely sections: 4,1,3,2 4, 1, 3, 2 4,1,3,2.3. Probabilities are calculated based on the number of favorable outcomes over the total number of outcomes.

STEP 2

1. Calculate the probability of landing on specific numbers for both spinners.2. Evaluate each statement based on the calculated probabilities.

STEP 3

Calculate the probability that spinner A A A lands on 1.Spinner A A A has two sections labeled 1 out of six total sections.P(A lands on 1)=26=13 P(A \text{ lands on } 1) = \frac{2}{6} = \frac{1}{3} P(A lands on 1)=62​=31​

STEP 4

Calculate the probability that spinner B B B lands on 1.Spinner B B B has one section labeled 1 out of four total sections.P(B lands on 1)=14 P(B \text{ lands on } 1) = \frac{1}{4} P(B lands on 1)=41​

STEP 5

Compare the probabilities for statement (A).13 is not twice 14 \frac{1}{3} \text{ is not twice } \frac{1}{4} 31​ is not twice 41​Statement (A) is false.

STEP 6

Calculate the probability that spinner B B B lands on an odd number.Odd numbers on spinner B B B are 1 and 3. There are two odd sections out of four.P(B lands on odd)=24=12 P(B \text{ lands on odd}) = \frac{2}{4} = \frac{1}{2} P(B lands on odd)=42​=21​Statement (B) is true.

STEP 7

STEP 8

Determine if spinner A A A is fair.A fair spinner would have equal probabilities for each section, which is not the case here.Spinner A A A is not fair.

STEP 9

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1. Spinner $A$ has six equally likely sections: $2, 1, 1, 3, 2, 3$ .
2. Spinner $B$ has four equally likely sections: $4, 1, 3, 2$ .
3. Probabilities are calculated based on the number of favorable outcomes over the total number of outcomes.

1. Calculate the probability of landing on specific numbers for both spinners.
2. Evaluate each statement based on the calculated probabilities.

Calculate the probability that spinner $A$ lands on 1.
Spinner $A$ has two sections labeled 1 out of six total sections.
$P(A \text{ lands on } 1) = \frac{2}{6} = \frac{1}{3}$

Calculate the probability that spinner $B$ lands on 1.
Spinner $B$ has one section labeled 1 out of four total sections.
$P(B \text{ lands on } 1) = \frac{1}{4}$

Compare the probabilities for statement (A).
$\frac{1}{3} \text{ is not twice } \frac{1}{4}$
Statement (A) is false.

Calculate the probability that spinner $B$ lands on an odd number.
Odd numbers on spinner $B$ are 1 and 3. There are two odd sections out of four.
$P(B \text{ lands on odd}) = \frac{2}{4} = \frac{1}{2}$
Statement (B) is true.

Determine if spinner $A$ is fair.
A fair spinner would have equal probabilities for each section, which is not the case here.
Spinner $A$ is not fair.