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You pick a card at random. Without putting the first card back, you pick a second card at random. What is the probability of picking an $8$ and then picking a $9$ ? Write your answer as a fraction or whole number.. $\newline$ ______

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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. What is the probability of picking an

8

and then picking a

9

? Write your answer as a fraction or whole number..

\newline

______

Determine Probability of Picking $8$ : First, we need to determine the probability of picking an $8$ from a standard deck of $52$ cards. There are four $8$ s in a deck (one for each suit: hearts, diamonds, clubs, and spades). The probability of picking an $8$ is the number of $8$ s divided by the total number of cards. $\newline$ P(Picking an $8$ ) = Number of $8$ s / Total number of cards $\newline$ P(Picking an $8$ ) = $\frac{4}{52}$
Calculate Probability of Picking $9$ After $8$ : Next, we need to calculate the probability of picking a $9$ after having picked an $8$ , without putting the $8$ back into the deck. Now there are $51$ cards left in the deck, and there are still four $9$ s (since we haven't picked any $9$ s yet). The probability of picking a $9$ from the remaining cards is the number of $9$ s divided by the remaining number of cards. $\newline$ $P(\text{Picking a } 9 \text{ after an } 8) = \frac{\text{Number of } 9\text{s}}{\text{Remaining number of cards}}$ $\newline$ $P(\text{Picking a } 9 \text{ after an } 8) = \frac{4}{51}$
Find Combined Probability of Both Events: Now, we need to find the combined probability of both events happening one after the other. This is found by multiplying the probabilities of each individual event. $\newline$ $P(\text{Picking an 8 and then a 9}) = P(\text{Picking an 8}) \times P(\text{Picking a 9 after an 8})$ $\newline$ $P(\text{Picking an 8 and then a 9}) = (\frac{4}{52}) \times (\frac{4}{51})$
Perform Multiplication to Find Probability: Finally, we perform the multiplication to find the combined probability. $\newline$ $P(\text{Picking an 8 and then a 9}) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652}$ $\newline$ To simplify the fraction, we can divide both the numerator and the denominator by the greatest common divisor, which is $4$ . $\newline$ $P(\text{Picking an 8 and then a 9}) = \frac{16}{4} / \frac{2652}{4} = \frac{4}{663}$

More problems from Probability of independent and dependent events

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college students running for student government class president. The candidates include

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, the probability that event

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, and the probability that events

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B

\frac{1}{9}

\newline

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B

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Question

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Question

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Question

There are two different raffles you can enter.

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\newline

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\newline

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\newline

\text{[B]Raffle B}

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Question

X

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and standard deviation

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\newline

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?

\newline

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\newline

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\newline

What is the probability that

X

62

68

?

\newline

Write your answer as a decimal rounded to the nearest thousandth.

\newline

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Question

Complete the statement. Round your answer to the nearest thousandth.

\newline

In a population that is normally distributed with mean

43

and standard deviation

3

30\%

of the values are those less than ______.

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