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Spinner $R$ determines the digit in the tens place, and Spinner $T$ determines the digit in the ones place. What is the probability that the two-digit number determined by spinning each spinner one time is an even number? $\newline$ A. $\frac{1}{2}$ $\newline$ B. $\frac{1}{4}$ $\newline$ C. $\frac{5}{16}$ $\newline$ D. $\frac{1}{8}$

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Find probabilities using the addition rule

Full solution

Q. Spinner

R

determines the digit in the tens place, and Spinner

T

determines the digit in the ones place. What is the probability that the two-digit number determined by spinning each spinner one time is an even number?

\newline

A.

\frac{1}{2}

\newline

B.

\frac{1}{4}

\newline

C.

\frac{5}{16}

\newline

D.

\frac{1}{8}

Define Events: Let's define the events: $\newline$ - Spinner $R$ determines the tens place. $\newline$ - Spinner $T$ determines the ones place. $\newline$ A number is even if its ones place is an even digit ( $0$ , $2$ , $4$ , $6$ , or $8$ ).
Find Probability: We need to find the probability that Spinner T lands on an even number since the tens place does not affect whether a number is even or odd. $\newline$ Assuming Spinner T has digits $0-9$ (as is typical for a spinner), there are $5$ even digits out of $10$ possible digits. $\newline$ So, the probability that Spinner T lands on an even number is $\frac{5}{10}$ , which simplifies to $\frac{1}{2}$ .
Consider Outcome: Since the outcome of Spinner $R$ does not affect the evenness of the two-digit number, we only need to consider the outcome of Spinner $T$ . $\newline$ Therefore, the probability that the two-digit number is even is the same as the probability that Spinner $T$ lands on an even number.

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