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Something to brush up your basics...yup am revising prob. today
Q1. If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces?
Q2. If someone draws a card at random from a deck and then replaces the first card, draws a second card, what is the probability that both cards will be aces?
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Something to brush up your basics...yup am revising prob. today
Q1. If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces?
P(Ace on 1st draw) = 4/52 = 1/13 P(Ace on 2nd draw) = 3/51 = 1/17 (without replacing)
Hence answer = 1/13 * 1/17 = 1/221
Q2. If someone draws a card at random from a deck and then replaces the first card, draws a second card, what is the probability that both cards will be aces?
Show more
P(Ace on 1st draw) = 4/52 = 1/13
P(Ace on 2nd draw) = 3/52 = 1/13 (with replacement)
Something to brush up your basics...yup am revising prob. today
Q1. If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces?
Q2. If someone draws a card at random from a deck and then replaces the first card, draws a second card, what is the probability that both cards will be aces?
Show more
As a non card player, are we expected to memorize that there are 52 cards in a deck for such probability problems? Or will that info be provided??
I have not seen a problem in which the number of cards in a deck is not provided, but I think its safe to assume that there are 52 cards in a pack, 13 of each suite, 26 of each color in a pack. Kinda like assuming there are 6 faces on a die.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.