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Each time she throws a dart, the probability that Mary hits the dartboard is
(2)/(7).
She throws two darts, one after the other.
What is the probability that she hits the dartboard with both darts?
(A)
(1)/(21)
(B)
(4)/(49)
(C)
(2)/(7)
(D)
(4)/(7)

Each time she throws a dart, the probability that Mary hits the dartboard is $\newline$ $\frac{2}{7}$ . $\newline$ She throws two darts, one after the other. $\newline$ What is the probability that she hits the dartboard with both darts? $\newline$ (A) $\frac{1}{21}$ $\newline$ (B) $\frac{4}{49}$ $\newline$ (C) $\frac{2}{7}$ $\newline$ (D) $\frac{4}{7}$

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Find probabilities using the addition rule

Full solution

Q. Each time she throws a dart, the probability that Mary hits the dartboard is

\newline

\frac{2}{7}

.

\newline

She throws two darts, one after the other.

\newline

What is the probability that she hits the dartboard with both darts?

\newline

(A)

\frac{1}{21}

\newline

(B)

\frac{4}{49}

\newline

(C)

\frac{2}{7}

\newline

(D)

\frac{4}{7}

Multiply probabilities: Since the events are independent, multiply the probability of hitting the dartboard on the first throw by the probability of hitting it on the second throw. $\newline$ $P(\text{both darts hit}) = P(\text{first dart hits}) \times P(\text{second dart hits})$
Substitute given probabilities: Substitute the given probability for each dart. $\newline$ $P(\text{both darts hit}) = \left(\frac{2}{7}\right) \times \left(\frac{2}{7}\right)$
Calculate product: Calculate the product. $P(\text{both darts hit}) = \frac{4}{49}$
Check answer choices: Check the answer choices to match the calculated probability. $\newline$ The correct answer is (B) $\frac{4}{49}$ .

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