Wiki User
∙ 12y ago20 out of 52.
Wiki User
∙ 12y ago5/13
2,4,6,8,10... 5/26
The probability of drawing a red even card is 5 in 13 (assuming that the face cards are neither even nor odd). The probability of throwing a sum of 11 is 1 in 18. The probability, then, of doing both of these actions, since they are mutually independent, is simply their product, or (5 in 13) times (1 in 18) equals 5 in 234, or about 0.02137.
The absolute probability is even, given one draw. However, statistically, the chance of drawing an ace and a king as two cards at random is 1: 81.25Chance of drawing first card is either an ace or a king is 8: 52 (1 in 6.5).Now the remaining other card (ace or king) is 4 in 51 (1 in 12.5)*In Blackjack, the drawing of any face card or 10 improves the odds of a natural blackjack using only one deck to 1: 20.8 but the show used holds more than one deck.
It's used commonly to estimate the traits of a child of two parents. For example, the probability of the child having blue eyes, or curly hair, or even having genetic disease.
5/13
The probability of selecting a red card is 26 in 52 or 1 in 2. The probability of selecting an even card is 20 in 52 or 5 in 13. The probability, therefore, of selecting a red even card is 1 in 2 times 5 in 13 or 5 in 26.
2,4,6,8,10... 5/26
It is 20/52 or 5/13.
The probability of drawing a red even card is 5 in 13 (assuming that the face cards are neither even nor odd). The probability of throwing a sum of 11 is 1 in 18. The probability, then, of doing both of these actions, since they are mutually independent, is simply their product, or (5 in 13) times (1 in 18) equals 5 in 234, or about 0.02137.
The question asks for the probability of an even card OR a red card. The term "OR" is key since this is not the same as the probability of drawing an even card and a red card, that is to say an even red card. GIven any two events, A and B P(A or B)=P(A)+P(B)-P(A and B) IF A and B are mutually exclusive, then P(A and B)=0 and this equation becomes P(A)+P(B) However, they are NOT in this case. So let A be the probability the card is even and B the probability it is red. P(A)=20/52 since J, K and Q are neither even nor odd (20=(52-12)/2)) P(B)=26/52 since half the cards are red. P(A and B) is the probability that a card is red AND even. We have 20 even cards, half of them are red and half are black so the odds are 10/52 of being red and even. P (A or B)=20/52+26/52-10/52=9/13
The odds of any card pulled from an ordinary deck of 52 cards being an Ace is 4 in 52 (4 aces in a deck of 52). This can be reduced to a 1 in 13 chance of any random card pulled from the deck being an Ace (or any other specific value, for that matter). That 13th last card dealt in a hand is no different than picking a random card out of the pack, regardless of what cards you deal before (face down or blindfolded or even face up, it doesn't matter). A more interesting question would be "what would the probability be of ANY of those 13 cards being an Ace?" Any takers?
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)
A standard 52-card deck would have five even-numbered hearts: 2, 4, 6, 8, and 10.
The absolute probability is even, given one draw. However, statistically, the chance of drawing an ace and a king as two cards at random is 1: 81.25Chance of drawing first card is either an ace or a king is 8: 52 (1 in 6.5).Now the remaining other card (ace or king) is 4 in 51 (1 in 12.5)*In Blackjack, the drawing of any face card or 10 improves the odds of a natural blackjack using only one deck to 1: 20.8 but the show used holds more than one deck.
The probability, or probility, even, is 0 since tere can be no such thing as "choosing red card of the black".
It's used commonly to estimate the traits of a child of two parents. For example, the probability of the child having blue eyes, or curly hair, or even having genetic disease.