更新1:
Never mind. There are not 5 of each type of card
Can someone help me with this math problem ?
2019-11-08 10:09 pm
A board game uses a deck of 20 cards. There are 5 of each type of card. Two cards are selected at random from this deck. Determine the probability that neither card shows a 2 or 3 both with and without replacement
回答 (2)
2019-11-08 10:33 pm
What if the card types are: yellow, dog, New Zealand, car, and water? Then the probability of showing a 2 or 3 is just about zero.
2019-11-08 10:20 pm
Assuming that "a 2" and "a 3" are types of cards, so that there are five of each of them, and ten other cards:
You need the first card to not be a 2 or 3 *and* the second card to not be a 2 or 3.
With replacement, the probabilities are the same for both cards Each choice has a 10/20 = 1/2 probability of not being a 2 or 3. Combined, that's a probability of (1/2)(1/2) = 1/4 of getting no 2 or 3 on both cards. That's exactly 0.25 as a decimal fraction.
Without replacement the first card has a 1/2 probability of not being a 2 or 3 (same as before) but if that succeeds, then there are only 19 cards left for the second draw, and only 9 of those aren't a 2 or 3. The combined probability is (1/2)(9/19) = 9/38, or approximately 0.237 as a decimal fraction.
You need the first card to not be a 2 or 3 *and* the second card to not be a 2 or 3.
With replacement, the probabilities are the same for both cards Each choice has a 10/20 = 1/2 probability of not being a 2 or 3. Combined, that's a probability of (1/2)(1/2) = 1/4 of getting no 2 or 3 on both cards. That's exactly 0.25 as a decimal fraction.
Without replacement the first card has a 1/2 probability of not being a 2 or 3 (same as before) but if that succeeds, then there are only 19 cards left for the second draw, and only 9 of those aren't a 2 or 3. The combined probability is (1/2)(9/19) = 9/38, or approximately 0.237 as a decimal fraction.
收錄日期: 2021-04-24 07:48:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20191108140914AAsUEqG