Bad form on this question. The answer is impossible to determine because we don't know if the first card was a heart or not. The question should have been phrased "The second card is red given that the first card is a heart". In this better scenario, there are 51 remaining cards, 25 of the remaining ones are red, so the answer is 25/51.
This is a conditional probability, given the card is red, what is the chance it is a heart. Since there are 2 red hearts, the probability if 1/2
It is 156/663 = 0.2353, approx.
Two cards are drawn from a pack of 52 cards second card is drawn after replacing the first card. What is the probability that the second card is a king?
Oh, dude, the probability of drawing 2 hearts from a deck of cards is like 1 out of 13 for the first card, and then 12 out of 51 for the second card. So, if you multiply those together, you get about a 4.5% chance of pulling off that heartwarming feat. But hey, who's counting, right?
First off, how do I calculate the probability that any one event occurs. The answer is equal to: Number of Possible Chances of Success / Total Number of Chances In this case, the number of possible chances of success is one (there is only one 6 of Diamonds in any deck of cards). The total number of chances equal 52 (there are 52 cards to choose from). Therefore the probability of picking a 6 of Diamonds on the first card is 1/52 or .019. In order to calculate the probability that the first card is a 6 of Diamonds AND the second card is a 3 of Hearts, you multiply the two probabilities. Prob. of 1st Card 6D AND 2nd Card 3H = Prob. 1st Card 6D * Prob. 2nd Card 3H We already know the probability of getting a 6 of Diamonds on the first card is 1/52 or .019. To calculate the probability of getting a 3 of Hearts on the second card, it is important to remember that random occurances do not affect the probability of other random occurrances. What I mean is, if I were to draw a 6 of Diamonds from a deck of cards and then replace it, the probability that I would pick a 6 of Diamonds again is the same as it was the first time. Even if I flip a coin 5 times in a row and they all landed on heads, the probability that I would flip another heads is still 50/50. So basically we can ignore what happened on the first draw, and jsut calculate the probability of getting a 3 of Hearts. Again we use our probability formula: Number of Possible Chances of Success / Total Number of Chances In this case, the number of possible chances of success is one (there is only one 3 of Hearts in any deck of cards). The total number of chances equal 52 (THIS ASSUMES THAT WE PUT THE 6 OF DIAMONDS BACK INTO THE DECK AFTER THE FIRST DRAW IF NOT THE NUMBER OF CHANCES IS 51). Therefore the probability of picking a 3 of Hearts on the second card is 1/52 or .019. Multiply the two probabilities together to get the probability of both occurring: 1/52 * 1/52 = 1/2704 = .00037 (or a .037 percent of a chance)
This is a conditional probability, given the card is red, what is the chance it is a heart. Since there are 2 red hearts, the probability if 1/2
It is 156/663 = 0.2353, approx.
Draw from the bottom?
The answer depends on whether or not the first card is replaced before the second is drawn.
The Jack of Hearts
Yes. In Hearts you have to throw a card of the same suit as the first card thrown if you have one. The first card to play in the game is always the two of clubs. If you did not have a club, you could throw a card of a different suit. However, you cannot throw a point card in the first round. This means you cannot throw a heart or the queen of spades. But the jack of diamonds would be legal.
pr(success) = number_of_ways_of_success/total_number_of_ways. As the two selections are independent, multiply the probability that the first card is a heart by the probability the second card is a heart. There are 13 hearts and 52 cards in a deck → pr(1st card heart) = 13/52 = 1/4 After 1 card has been selected, there are 51 cards left and if the first was a heart, there are only 12 hearts left → pr(2nd card heart also) = 12/51 = 4/17 → pr(1st two heart) = pr(1st card heart) × pr(2nd card heart) = 1/4 × 4/17 = 1/17
Two cards are drawn from a pack of 52 cards second card is drawn after replacing the first card. What is the probability that the second card is a king?
Since half of the cards (26 of the 52) are a red card (hearts and diamonds) the odds are 1 in 2 (or 50%) that any given card will be red.
'Hearts'.
If you mean when you say second group of offenses the second kind of offense you have made in the same bout, it should be a red card, if you have already collected a red or a yellow card. If you mean when you say second group of offenses a group of offenses you make in a different bout, it would be a yellow card, as red and yellow cards do not carry over to other bouts.
Just a straight deal, the first card will be an ace 4 times out of 52, the second card will be an ace 3 times in 51. The total odds will be 12 times in 2652.However in the question posed the first card is an ace so the probability is 1 in 1, so the second card being an ace will occur 3 times in 51.In the question , the first card is ACE, and second is also an ACE. Hence probability is 1in 1.How ever the question should be framed "What is the probabilty for getting first card An ACE and also Second card as An ACE"