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
For the first card, you have 13 (favorable) options out of 52 (13/52), since the deck of cards has 13 hearts. For the second card, you have 12 options out of 51 (12/51), since there is one card less, and one heart has already been taken. Multiply those two fractions.
It is 1/17.
If you are drawing two cards from a full deck of cards (without jokers) then the probability will depend upon whether the the first card is replaced before the second is drawn, but the probability will also be different to being dealt a hand whilst playing Bridge (or Whist), which will again be different to being dealt a hand at Canasta. Without the SPECIFIC context of the two cards being got, I cannot give you a more specific answer.
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The Binomial Probability DistributionA binomial experiment is one that possesses the following properties:On this page...Mean and variance of a binomial distributionThe experiment consists of n repeated trials;Each trial results in an outcome that may be classified as a success or a failure (hence the name, binomial);The probability of a success, denoted by p, remains constant from trial to trial and repeated trials are independent.The number of successes X in n trials of a binomial experiment is called a binomial random variable.The probability distribution of the random variable X is called a binomial distribution, and is given by the formula:P(X) = Cnxpxqn−xwheren = the number of trialsx = 0, 1, 2, ... np = the probability of success in a single trialq = the probability of failure in a single trial(i.e. q = 1 − p)Cnx is a combinationP(X) gives the probability of successes in n binomial trials.Mean and Variance of Binomial DistributionIf p is the probability of success and q is the probability of failure in a binomial trial, then the expected number of successes in n trials (i.e. the mean value of the binomial distribution) isE(X) = μ = npThe variance of the binomial distribution isV(X) = σ2 = npqNote: In a binomial distribution, only 2 parameters, namely n and p, are needed to determine the probability.EXAMPLE 1Image sourceA die is tossed 3 times. What is the probability of(a) No fives turning up?(b) 1 five?(c) 3 fives?AnswerLoading...EXAMPLE 2Hospital records show that of patients suffering from a certain disease, 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?AnswerLoading...EXAMPLE 3Image sourceIn the old days, there was a probability of 0.8 of success in any attempt to make a telephone call.Calculate the probability of having 7 successes in 10 attempts.AnswerLoading...EXAMPLE 4A (blindfolded) marksman finds that on the average he hits the target 4 times out of 5. If he fires 4 shots, what is the probability of(a) more than 2 hits?(b) at least 3 misses?AnswerLoading...EXAMPLE 5Image sourceThe ratio of boys to girls at birth in Singapore is quite high at 1.09:1.What proportion of Singapore families with exactly 6 children will have at least 3 boys? (Ignore the probability of multiple births.)[Interesting and disturbing trivia: In most countries the ratio of boys to girls is about 1.04:1, but in China it is 1.15:1.]AnswerLoading...EXAMPLE 6A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain(a) no more than 2 rejects? (b) at least 2 rejects?AnswerLoading...11. Probability Distributions - Concepts13. Poisson Probability DistributionDidn't find what you are looking for on this page? Try search:The IntMath NewsletterSign up for the free IntMath Newsletter. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!Given name: * requiredFamily name:email: * requiredSee the Interactive Mathematics spam guarantee.Probability Lessons on DVDEasy to understand probability lessons on DVD. See samples before you commit.More info: Probability videosBookmark this pageAdd this page to diigo, Redditt, etc.Need a break? Play a math game. Well, they all involve math... No, really!Help keep Interactive Mathematics free!Home | Sitemap | About & Contact | Feedback & questions | Privacy | IntMath feed |Hello, PakistanPage last modified: 22 March 2007Valid HTML 4.01 | Valid CSSChapter ContentsCounting and Probability - Introduction1. Factorial Notation2. Basic Principles of Counting3. Permutations4. Combinations5. Introduction to Probability Theory6. Probability of an EventSingapore TOTOProbability and Poker7. Conditional Probability8. Independent and Dependent Events9. Mutually Exclusive Events10. Bayes' Theorem11. Probability Distributions - Concepts12. Binomial Probability Distributions13. Poisson Probability Distribution14. Normal Probability DistributionThe z-TableFollowing are the original SNB files (.tex or .rap) used in making this chapter. For more information, go to SNB info. SNB files1. Factorial Notation (SNB)2. Basic Principles of Counting (SNB)3. Permutations (SNB)4. Combinations (SNB)5. Introduction to Probability Theory (SNB)6. Probability of an Event (SNB)Singapore TOTO (SNB)Probability and Poker (SNB)7. Conditional Probability (SNB)8. Independent and Dependent Events (SNB)9. Mutually Exclusive Events (SNB)10. Bayes' Theorem (SNB)11. Probability Distributions - Concepts (SNB)12. Binomial Probability Distributions (SNB)13. Poisson Probability Distribution (SNB)14. Normal Probability Distribution (SNB)Comments, Questions?Math ApplicationsI get a good understanding of how math is applied to real world problems:In most lessonsIn some lessonsRarelyNeverVotes so far: 1570Follow IntMath on TwitterGet the Daily Math Tweet!IntMath on TwitterRecommendationEasy to understand probability lessons on DVD. Try before you commit. 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No, they don't ---- It depends on what sport they are playing. If they are playing cricket, football etc then they would find it dufficult playing without a ball
~18102 if you assume that each playing card is 3.5 inches in length.
playing
If you select 45 cards without replacement from a regular deck of playing cards, the probability is 1. For a single randomly selected card, the probability is 2/13.
"Playing cards" are chosen at random.
If the card is drawn randomly, the probability is 1/4.
The answer depends onwhether the card(s) are drawn from a normal deck of playing cards,whether they are at random,how many cards are drawn,whether the cards are replaced before drawing the next card.Thus, if 49 cards are drawn without replacement from an ordinary deck, whether randomly or not, the probability is 1.For a single card drawn randomly, the probability is 1/13.The answer depends onwhether the card(s) are drawn from a normal deck of playing cards,whether they are at random,how many cards are drawn,whether the cards are replaced before drawing the next card.Thus, if 49 cards are drawn without replacement from an ordinary deck, whether randomly or not, the probability is 1.For a single card drawn randomly, the probability is 1/13.The answer depends onwhether the card(s) are drawn from a normal deck of playing cards,whether they are at random,how many cards are drawn,whether the cards are replaced before drawing the next card.Thus, if 49 cards are drawn without replacement from an ordinary deck, whether randomly or not, the probability is 1.For a single card drawn randomly, the probability is 1/13.The answer depends onwhether the card(s) are drawn from a normal deck of playing cards,whether they are at random,how many cards are drawn,whether the cards are replaced before drawing the next card.Thus, if 49 cards are drawn without replacement from an ordinary deck, whether randomly or not, the probability is 1.For a single card drawn randomly, the probability is 1/13.
A card is drawn from a standard deck of playing cards. what is the probability that a spade and a heart is selected?
The probability is 0.25
The answer depends on how many cards are drawn, whether or not at random, with or without replacement. The probability for a single card, drawn at random, from a normal deck of playing cards is 2/13.The answer depends on how many cards are drawn, whether or not at random, with or without replacement. The probability for a single card, drawn at random, from a normal deck of playing cards is 2/13.The answer depends on how many cards are drawn, whether or not at random, with or without replacement. The probability for a single card, drawn at random, from a normal deck of playing cards is 2/13.The answer depends on how many cards are drawn, whether or not at random, with or without replacement. The probability for a single card, drawn at random, from a normal deck of playing cards is 2/13.
You randomly select one card from a 52-card deck. Find the probability of selecting the king of diamonds or the jack of
The probability is 0.
The probability is 1/13 of drawing a king in one draw from a standard deck with no jokers.
The probability is 16/52 = 4/13
The probability of drawing an Ace from a standard deck of 52 cards is 4 in 52, or 1 in 13.