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You pick a card at random. Without putting the first card back, you pick a second card at random. $3$ $4$ $5$ $6$ $7$ $8$ $9$ What is the probability of picking a $5$ and then picking an even number? Write your answer as a fraction or whole number.

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Probability of independent and dependent events

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Q. You pick a card at random. Without putting the first card back, you pick a second card at random.

3

4

5

6

7

8

9

What is the probability of picking a

5

and then picking an even number? Write your answer as a fraction or whole number.

Determine Probability of $5$ : First, we need to determine the probability of picking a $5$ from the sequence. $\newline$ There are $7$ cards in total, and only one of them is a $5$ . $\newline$ So, the probability of picking a $5$ is $1$ out of $7$ . $\newline$ Calculation: $P(\text{picking a } 5) = \frac{1}{7}$
Probability of Even Number After $5$ : Next, we need to determine the probability of picking an even number after having picked a $5$ . After picking a $5$ , there are $6$ cards left. Out of these $6$ cards, there are $3$ even numbers ( $4$ , $6$ , $8$ ). So, the probability of picking an even number after a $5$ is $3$ out of $6$ . Calculation: $5$ $1$
Multiply Probabilities: Now, we need to multiply the two probabilities together to find the overall probability of both events happening in sequence. $\newline$ Calculation: $P(\text{picking a } 5 \text{ and then an even number}) = P(\text{picking a } 5) \times P(\text{picking an even number after a } 5) = \frac{1}{7} \times \frac{3}{6}$
Final Probability Calculation: Finally, we perform the multiplication to get the final probability. $\newline$ Calculation: $(\frac{1}{7}) \times (\frac{3}{6}) = \frac{3}{42}$ $\newline$ This fraction can be simplified by dividing both the numerator and the denominator by $3$ . $\newline$ Calculation: $\frac{3}{42} = \frac{1}{14}$

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