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You choose a card at random. Without putting the first card back, you choose a second card at random. What is the probability of choosing a $10$ and then choosing a Jack? 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 choose a card at random. Without putting the first card back, you choose a second card at random. What is the probability of choosing a

10

and then choosing a Jack? Write your answer as a fraction or whole number.

Determine total ways: First, we need to determine the total number of ways to choose any card from a deck. A standard deck of cards has $52$ cards. So, the total number of ways to choose the first card is $52$ .
Calculate $10$ probability: Next, we calculate the probability of choosing a $10$ as the first card. There are four $10$ s in a deck (one for each suit), so the probability of choosing a $10$ on the first draw is the number of $10$ s divided by the total number of cards. $\newline$ $P(\text{Choosing a 10}) = \frac{\text{Number of 10s}}{\text{Total number of cards}} = \frac{4}{52}$
Calculate Jack probability: After choosing a $10$ , we do not put it back in the deck, so now there are $51$ cards left. We need to calculate the probability of choosing a Jack from the remaining cards.
Find overall probability: There are four Jacks in a deck. So, the probability of choosing a Jack after a $10$ has been chosen is the number of Jacks divided by the remaining number of cards. $P(\text{Choosing a Jack after a } 10) = \frac{\text{Number of Jacks}}{\text{Remaining number of cards}} = \frac{4}{51}$
Perform multiplication: To find the overall probability of both events happening in sequence (choosing a $10$ and then a Jack), we multiply the probabilities of each individual event. $\newline$ $P(\text{Choosing a } 10 \text{ and then a Jack}) = P(\text{Choosing a } 10) \times P(\text{Choosing a Jack after a } 10) = \left(\frac{4}{52}\right) \times \left(\frac{4}{51}\right)$
Find probability: Now we perform the multiplication to find the probability. $P(\text{Choosing a 10 and then a Jack}) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652}$
Simplify fraction: We can simplify the fraction by dividing both the numerator and the denominator by the greatest common divisor, which is $4$ . $P(\text{Choosing a } 10 \text{ and then a Jack}) = \frac{16}{4} / \frac{2652}{4} = \frac{4}{663}$
Final probability: The simplified probability of choosing a $10$ and then a Jack without replacement is $\frac{4}{663}$ .

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