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What is the probability of choosing a king on the second draw if the first was a king without replacement?
There are 4 Kings in a standard pack of 52 cards. If 1 King has previously been drawn this now leaves 3 kings out of a total of 51 remaining cards. The probability of now drawing a King is therefore 3/51 which simplifies to 1/17. Note: this is the probability concerning the 2nd draw only.
When a bag contains 4 white chips and 6 black chips. What is the probability of randomly choosing a white chip not replacing it and then randomly choosing another white chip?
To find the probability of choosing two white chips in succession without replacement, we first calculate the probability of selecting a white chip on the first draw. There are 4 white chips out of a total of 10 chips, so the probability of the first draw is 4/10. After removing one white chip, there are 3 white chips left out of a total of 9 chips, making the probability of the second draw 3/9. Therefore, the overall probability of drawing two white chips in succession is (4/10) * (3/9) = 12/90, which simplifies to 2/15.
If 2 cards are selected from a standard deck of 52 cards without replacement find the probability that both are the same suit?
If 2 cards are selected from a standard deck of 52 cards without replacement, in order to find the probability that both are the same suit, start with the first card...The probability that the first card is any suit is 52 in 52, or 1.Now, consider the second card. There are 12 cards remaining in the same suit, and 39 cards remaining in the other three suits...The probability that the second card is the same suit as the first card is 12 in 51, or 4 in 17, or 0.235.The probability of both events occurring is the product of those two probabilities. That is still 4 in 17, or 0.235.
Three cards are chosen at random from a standard deck of cards without replacement What is the probability of getting 3 aces?
The probability of drawing the first ace is 4 in 52. The probability of getting the second ace is 3 in 51. The probability of getting the third ace is 2 in 50. The probability, then, of drawing three aces is (4 in 52) times (3 in 51) times (2 in 50), which is 24 in 132600, or 1 in 5525, or about 0.0001810
What is the probability of selcting two s's if the first card is not replaced before selcting the second card?
There are no s's in a standard deck of cards, so the probability of selecting any s's, in any sequence of draws, in any strategy of replacement is exactly zero.
What is the probability of choosing a face card of a jack or queen or king on the second draw if the first draw was a ace without replacement?
The probability of drawing a jack, queen, or king on the second draw if the first draw was an ace (without replacement) is (4 + 4 + 4) in (52 - 1) or 12 in 51, which is 4 in 17, or about 0.2353.
What is the probability of choosing a king on the second draw if the first was a king without replacement?
There are 4 Kings in a standard pack of 52 cards. If 1 King has previously been drawn this now leaves 3 kings out of a total of 51 remaining cards. The probability of now drawing a King is therefore 3/51 which simplifies to 1/17. Note: this is the probability concerning the 2nd draw only.
When a bag contains 4 white chips and 6 black chips. What is the probability of randomly choosing a white chip not replacing it and then randomly choosing another white chip?
To find the probability of choosing two white chips in succession without replacement, we first calculate the probability of selecting a white chip on the first draw. There are 4 white chips out of a total of 10 chips, so the probability of the first draw is 4/10. After removing one white chip, there are 3 white chips left out of a total of 9 chips, making the probability of the second draw 3/9. Therefore, the overall probability of drawing two white chips in succession is (4/10) * (3/9) = 12/90, which simplifies to 2/15.
What is the probability of choosing a face card on the second draw if the first draw was a king?
7/26
If you have a bag of 4 red and 4 green marbles what is the probability of drawing 2 marbles without replacement and getting 2 green marbles?
4/8 or 1/2(probability of first draw) * 3/8(probability of second draw which is 12/64 or 3/16 of the given scenario.
How do you solve Two cards are chosen at random from a deck of 52 cards without replacement What is the probability that the first card is a jack and the second card is a ten?
The probability that the first card is a jack is 4 in 52. The probability that the second card is 1 ten is 4 in 51. Since these are sequential events, simply multiply, giving (4/52)(4/51) or (16/2652) or about 0.00603.
Two cards are drawn from a deck of cards. what is the probability that both cards are spades?
This question is a little bit tricky. In a deck of 52 cards, one-fourth or 13 cards are spades. So, the chance of drawing one spade = 13/52 or 0.25. If a second card drawn, there's one less spade in the deck, so the probability on the second draw is 12/51. The probability of drawing two spades from a deck is 0.25 x 12/51 = 0.058824 This is called sampling without replacement. In quality control, it is very common to sample without replacement as bad parts are discarded. If we consider drawing one card, putting it back in the deck, regardless if it is a spade or not, then reshuffling the deck and drawing the second card, the probability is 0.25 x 0.25 = 0.0625, a bit higher with replacement. This is the same as 1/4 x 1/4 = 1/8 or saying the odds are 1:8. I've included a couple of links on sampling with replacement and without replacement. Generally, for calculating statistics, we attempt to get independent results. The draw of one card, will reduce the population, and change the probabilities on the second draw, so sampling without replacement is not independent sampling. See related links.
If 2 cards are selected from a standard deck of 52 cards without replacement find the probability that both are the same suit?
If 2 cards are selected from a standard deck of 52 cards without replacement, in order to find the probability that both are the same suit, start with the first card...The probability that the first card is any suit is 52 in 52, or 1.Now, consider the second card. There are 12 cards remaining in the same suit, and 39 cards remaining in the other three suits...The probability that the second card is the same suit as the first card is 12 in 51, or 4 in 17, or 0.235.The probability of both events occurring is the product of those two probabilities. That is still 4 in 17, or 0.235.
What is the probability of getting 3 red balls from an urn in 3 picks without replacement if the urn contains 5 red 3 blue?
The probability the first ball will be red is 5/8. The probability that the first and second balls will be red is 5/8 x 4/7. The probability that the first, second, and third balls will be red is 5/8 x 4/7 x 3/6, or overall 60/336 (about 17.86%).
You select a card then you select again without replacement is it independent or dependent?
Selecting a card and then selecting again without replacement is a dependent event. This is because the outcome of the second selection is influenced by the result of the first selection; the total number of cards decreases and potentially alters the probabilities of the remaining cards. Thus, the events are not independent, as the probability of selecting a specific card in the second draw depends on what was drawn in the first.
Three cards are chosen at random from a standard deck of cards without replacement What is the probability of getting 3 aces?
The probability of drawing the first ace is 4 in 52. The probability of getting the second ace is 3 in 51. The probability of getting the third ace is 2 in 50. The probability, then, of drawing three aces is (4 in 52) times (3 in 51) times (2 in 50), which is 24 in 132600, or 1 in 5525, or about 0.0001810
Draw two cards from a deck of cards without replacement what is the probability that the first card is an ace and the second card is a king?
The probability that two cards drawn from a deck of cards being an Ace followed by a King is 1 in 13 (for the Ace) times 4 in 51 (for the King) which is equal to 4 in 663.
