What is the probability of drawing either hearts or diamond randomly from standard deck of cards

In such a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and diamonds are red. The 26 cards included in clubs and spades are black. The 13 cards in each suit are: 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, and Ace. This means there are four Aces, four Kings, four Queens, four 10s, etc., down to four 2s in each deck.You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second.(a) Are the outcomes on the two cards independent? Why?No. The events cannot occur together.No. The probability of drawing a specific second card depends on the identity of the first card.    Yes. The probability of drawing a specific second card is the same regardless of the identity of the first drawn card.Yes. The events can occur together.(b) Find P(ace on 1st card and nine on 2nd). (Enter your answer as a fraction.)(c) Find P(nine on 1st card and ace on 2nd). (Enter your answer as a fraction.)(d) Find the probability of drawing an ace and a nine in either order. (Enter your answer as a fraction.)

What is the probability of drawing either hearts or diamond randomly from standard deck of cards

The probability of drawing a diamond-faced card from a pack of 52 playing cards is easy to determine. Since there are 13 diamond-faced cards in the deck, the probability becomes 13/52 = 1/4 = 0.25.

The probability of drawing an ace from a pack of 52 playing cards is also easy to determine. There are 4 aces in the deck of 52 cards; thus, the probability becomes 4/52 = 1/13 = 0.076923. This represents a much lower probability than drawing a card in a specific suit, illustrated in the preceding example.

A branch of mathematics that deals with the happening of a random event is termed probability. It is used in Maths to predict how likely events are to happen. The probability of any event can only be between 0 and 1 and it can also be written in the form of a percentage.

Probability

The probability of event A is generally written as P(A). Here, P represents the possibility and A represents the event. It states how likely an event is about to happen. The probability of an event can exist only between 0 and 1 where 0 indicates that event is not going to happen i.e. Impossibility and 1 indicates that it is going to happen for sure i.e. Certainty.

If not sure about the outcome of an event, take help of the probabilities of certain outcomes, how likely they occur. For a proper understanding of probability, take an example of tossing a coin, there will be two possible outcomes – heads or tails.

Formula of Probability

Probability of an event, P(A) = Favorable outcomes / Total number of outcomes

There are different terms used in the probability that is not commonly used normally, terms like experiments, sample space, a favorable outcome, trial, random experiment, etc. Let’s take a look at their definitions in detail,

  • Experiment: An operation or trial done to produce an outcome is called an experiment.
  • Sample Space: An experiment together constitutes a sample space for all the possible outcomes. For example, the sample space of tossing a coin is head and tail.
  • Favorable Outcome: An event that has produced the required result is called a favorable outcome. For example, If two dice are rolled at the same time then the possible or favorable outcomes of getting the sum of numbers on the two dice as 4 are (1, 3), (2, 2), and (3, 1).
  • Trial: A trial means doing a random experiment.
  • Random Experiment: A random experiment is an experiment that has a well-defined set of outcomes. For example, when a coin is tossed, a head or tail is obtained but the outcome is not sure which one will appear.
  • Event: An event is the outcome of a random experiment.
  • Equally Likely Events: Equally likely events are rare events that have the same chances or probability of occurring. Here The outcome of one event is independent of the other. For instance, when a coin is tossed, there are equal chances of getting a head or a tail.
  • Exhaustive Events: An exhaustive event is when the set of all outcomes of an experiment is equal to the sample space.
  • Mutually Exclusive Events: Events that cannot happen simultaneously are called mutually exclusive events. For example, the climate can be either cold or hot. One cannot experience the same weather again and again.

The Possibility of only two outcomes which is an event will occur or not, like a person will eat or not eat the food, buying a bike or not buying a bike, etc. are examples of complementary events.

Some Probability Formulae

  • Addition rule: Union of two events, say A and B, then,

P(A or B) = P(A) + P(B) – P(A∩B)

P(A∪ B) = P(A) + P(B) – P(A∩B)

  • Complementary rule: If there are two possible events of an experiment so the probability of one event will be the Complement of another event. For example, if A and B are two possible events, then,

P(B) = 1 – P(A) or P(A’) = 1 – P(A).

P(A) + P(A′) = 1.

  • Conditional rule: When the probability of an event is given and the second is required for which first is given, then P(B, given A) = P(A and B), P(A, given B). It can be vice versa,

P(B∣A) = P(A∩B)/P(A)

  • Multiplication rule: Intersection of two other events i.e. events A and B need to occur simultaneously. Then

P(A and B) = P(A)⋅P(B).

P(A∩B) = P(A)⋅P(B∣A)

Solution:

It is known that a well-shuffled deck has 52 cards

Total number of black cards = 26

Total number of red cards = 26

further divided into suits (4 of them: Spades, Hearts, Diamonds, Clubs) of 13 cards each. 

And Each suit has 13 cards (A, 2 to10, Jack, Queen, King).

So , total number of outcome = 52 

probability of getting either a heart or a jack?

probability of getting a heart = 13 

probability of getting a jack   = 4

And probability of getting a jack of heart = 1

Therefore probability of getting a heart = {total number of heart cards in the deck}/{total number of cards in the deck}

 = 13/52

Probability of getting a heart = 1/4

And the probability of getting either a jack = {total number of jack cards in the deck}/{total number of cards in the deck}

= 4/52

Probability of getting a jack = 1/13

 probability of getting a jack of heart = {total number of jack of heart in the deck}/{total number of cards in the deck}

= 1/52

Similar Problems

Question 1: What is the probability of getting a queen or a red card?

Solution:

Total number of cards are 52 

number of red cards are 26 and queens are 4 whereas 26 red cards contain 2 queens(so only 2 will be considered out of 4).

So, total outcomes = 52

favorable outcomes = 26 + 2 = 28

So, the probability of getting a queen or red card = Favorable outcomes / Total outcomes

= 28 / 52 = 7/13

P = 7/13

Question 2: What is the probability of drawing a black card from a well-shuffled deck of 52 cards?

Solution:

We know that a well-shuffled deck has 52 cards

Total number of black cards = 26

Total number of red cards = 26

Therefore probability of getting a black card= {total number of black cards in the deck}/{total number of cards in the deck}

= 26/52

= 1/2

So the probability of having black card is 1/2

Question 3:  What is the probability of getting a black queen or a diamond?

Solution:

Total number of cards = 52

Number  of favorable cards that are black queen = 2

so, probability of getting a black queen = 2/52  

Total number of cards that are diamond =13

Therefore probability of getting a diamond = {13/52}    

Therefore, probability of getting a black queen = 2/52

P(E) = probability of getting a black queen + probability of getting a diamond

= 2/52 +13/52

= 15/32

Question 4:  Find the probability of getting a number less than 5 in a single dice throw.

Solution:

When the dice is rolled then there will be 6 outcomes.

Total number of favorable outcome {set of outcome} = {1, 2, 3, 4, 5, 6}

= 6

Now as per the question,

Probability of getting a number less than 5 in a single throw is 4

Numbers less than 5 are {1,2,3,4}

therefore favorable outcome will be = 4

P(A) = Favorable outcomes / Total number of outcomes

= 4/6

= 2/3

Hence the probability of getting a number less than 5 in a single throw of a die is 2/3

Question 5: What are the odds of flipping 7 heads in a row?

Solution:

Probability of an event = (number of favorable event) / (total number of event).

P(B) = (occurrence of Event B) / (total number of event).

Probability of getting one head = 1/2.

here Tossing a coin is an independent event, its not dependent on how many times it has been tossed.

Probability of getting 2 heads in a row = probability of getting head first time × probability of getting head second time.

Probability of getting 2 head in a row = (1/2) × (1/2)

Therefore, the probability of flipping 7 heads in a row = (1/2)7

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