What is the probability of drawing a 10 from a deck of 52?

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Say we have a standard deck of 52 cards. Probability of drawing the King of Hearts OR the Queen of Diamonds is $\frac{2}{52}$ obviously (or $\frac{1}{26}$). If we were to make 30 draws with replacement (so each time the card is drawn, it is put back in the deck and and the deck is shuffled), the odds that the King of Hearts OR the Queen of Diamonds card was drawn at least once out of those 30 draws? And the answer is $1-\left((\frac{50}{52}\right)^{30})$

But now, what if we had the same situation but instead had asked: What are the odds that the King of Hearts OR the Queen of Diamonds card was drawn at least ten times out of those 30 draws?

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Probability means Possibility. 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. 
The higher or lesser the probability of an event, the more likely it is that the event will occur or not respectively. 

For example – An unbiased coin is tossed once. So the total number of outcomes can be 2 only i.e. either “heads” or “tails”. The probability of both outcomes is equal i.e. 50% or 1/2.

So, the probability of an event is Favorable outcomes/Total number of outcomes. It is denoted with the parenthesis i.e. P(Event).

P(Event) = N(Favorable Outcomes) / N (Total Outcomes)

Note: If the probability of occurring of an event A is 1/3 then the probability of not occurring of event A is 1-P(A) i.e. 1 – (1/3) = 2/3

What is Sample Space?

All the possible outcomes of an event are called Sample spaces.

Examples:

• A six faced dice is rolled once. So, total outcomes can be 6 and 
  Sample space will be [1, 2, 3, 4, 5, 6]
   
• An unbiased coin is tossed, So, total outcomes can be 2 and 
  Sample space will be [Head, Tail]
   
• If two dice are rolled together then total outcomes will be 36 and 
  Sample space will be [ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)   (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)   (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)   (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)   (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) 

    (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) ]

Types of Events

Independent Events: If two events (A and B) are independent then their probability will be P(A and B) = P (A ∩ B) = P(A).P(B) i.e. P(A) * P(B)

Example: If two coins are flipped, then the chance of both being tails is 1/2 * 1/2 = 1/4

Mutually exclusive events:

• If event A and event B can’t occur simultaneously, then they are called mutually exclusive events.
• If two events are mutually exclusive, then the probability of both occurring is denoted as P (A ∩ B) 
  and P (A and B) = P (A ∩ B) = 0
• If two events are mutually exclusive, then the probability of either occurring is denoted as P (A ∪ B) P (A or B) = P (A ∪ B)    = P (A) + P (B) − P (A ∩ B)    = P (A) + P (B) − 0  

    = P (A) + P (B)

Example – The chance of rolling a 2 or 3 on a six-faced die is P (2 or 3) = P (2) + P (3) = 1/6 + 1/6 = 1/3

Not Mutually exclusive events: If the events are not mutually exclusive then

P (A or B) = P (A ∪ B) = P (A) + P (B) − P (A and B)

What is Conditional Probability?

For the probability of some event A, the occurrence of some other event B is given. It is written as P (A ∣ B)

P (A ∣ B) = P (A ∩ B) / P (B)

Example- In a bag of 3 black balls and 2 yellow balls (5 balls in total), the probability of taking a black ball is 3/5, and to take a second ball, the probability of it being either a black ball or a yellow ball depends on the previously taken out ball. Since, if a black ball was taken, then the probability of picking a black ball again would be 1/4, since only 2 black and 2 yellow balls would have been remaining, if a yellow ball was taken previously, the probability of taking a black ball will be 3/4.

Some points related to Cards:

• There are 52 cards in a deck.
• In 52 cards, there are 26 cards of each color i.e. 26 red and 26 black cards.
• In 26 red cards, there are 2 suits of 13 cards each i.e. 13 heart and 13 diamond cards.
• In 26 black cards, there are 2 suits of 13 cards each i.e. 13 spades and 13 club cards.
• Each suite has 13 cards from 2 to 10, J, Q, K, and A which means 4 cards of each type.
• J, Q, and K are known as Face cards.

Answer:

Total number of cards are 52 and number of jack cards are 12 and number of 10 is 4.

So, total outcomes = 52 
Favorable outcomes = 4 + 4 = 8 (4-10s and 4-Js)

So, the probability of getting a 10 or a Jack = Favorable outcomes / Total outcomes = 8 / 52 = 2/13

P(10 or J) = 2/13

Similar Questions

Question 1: What is the probability of getting a jack or black card?

Solution:

Total number of cards are 52 and number of black cards are 26 and jacks are 4 whereas 26 black cards contain 2 Jacks(so only 2 will be considered out of 4)

So, total outcomes = 52
favorable outcomes = 26 + 2 = 28

So, the probability of getting a jack or black card = Favorable outcomes / Total outcomes = 28 / 52 = 7/13

P(JorB) = 7/13

Question 2: What is the probability of getting a queen or a card of diamonds?

Solution:

Total number of cards are 52 and number of diamonds cards are 13 and queens are 4 whereas 13 diamond cards contain 1 Q(so only 3 will be considered out of 4)

So, total outcomes = 52
favorable outcomes = 13 + 3 = 16

So, the probability of getting a queen or a diamond card = Favorable outcomes / Total outcomes = 16/ 52 = 4/13

P(QorD) = 4/13

Question 3: What is the probability of getting a jack or a red card?
Solution:

Total number of cards are 52 and number of red cards are 26 and jacks are 4 whereas 26 red cards contain 2 Jacks(so only 2 will be considered out of 4).

So, total outcomes = 52
favorable outcomes = 26 + 2 = 28

So, the probability of getting a jack or red card = Favorable outcomes / Total outcomes = 28 / 52 = 7/13

P(JorR) = 7/13

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