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$Independent and Dependent Events Suppose two cards are drawn randomly. What is the probability of drawing two green cards, if the first card is NOT replaced before the second draw? Assume the first card drawn is green. Show your answer as a fraction in lowest terms. Enter the numerator.$

Independent and Dependent Events $\newline$ Suppose two cards are drawn randomly. $\newline$ What is the probability of drawing two green cards, if the first card is NOT replaced before the second draw? Assume the first card drawn is green. $\newline$ Show your answer as a fraction in lowest terms. Enter the numerator.

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Independence and conditional probability

Full solution

Q. Independent and Dependent Events

\newline

Suppose two cards are drawn randomly.

\newline

What is the probability of drawing two green cards, if the first card is NOT replaced before the second draw? Assume the first card drawn is green.

\newline

Show your answer as a fraction in lowest terms. Enter the numerator.

Assume Green Cards in Deck: Let's assume there are $G$ green cards in the deck. When the first green card is drawn, there are now $G-1$ green cards left in a deck of one less card. The probability of drawing a green card on the first draw is $1$ , since we are given that the first card drawn is green. The probability of drawing another green card on the second draw is $(G-1)/(\text{Total cards in deck} - 1)$ . We need to know the total number of green cards and the total number of cards in the deck to calculate this probability.
Calculate Probability of Second Draw: Since we are not given the specific number of green cards or the total number of cards in the deck, we cannot calculate the exact probability. However, we can express the probability in terms of $G$ (the number of green cards) and $T$ (the total number of cards in the deck). The probability of drawing the first green card is $1$ (since it's given), and the probability of drawing the second green card is $\frac{G-1}{T-1}$ . Therefore, the combined probability is $1 \times \frac{G-1}{T-1} = \frac{G-1}{T-1}$ .
Express Probability in Terms of G and T: We realize that without the specific numbers for $G$ and $T$ , we cannot simplify this fraction further. The problem does not provide enough information to calculate a numerical probability. Therefore, we can only express the probability in terms of $G$ and $T$ as $\frac{G-1}{T-1}$ .

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\newline

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\newline

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