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.
- What is the probability of getting a queen and a jack card?
Solution:
Total number of cards are 52 and number of queens and jacks in 52 cards are 4 and 4 respectively.
So, total outcomes = 52
favorable outcomes = 4 + 4 = 8So, the probability of getting a queen or a jack = Favorable outcomes/Total outcomes = 8/52
P(J or Q) = 2/13
Similar Questions
Question 1: What is the probability of getting a queen of black card?
Solution:
Total number of cards are 52 and number of black queen card in 52 cards are 2.
So, total outcomes = 52
favorable outcomes = 2So, the probability of getting a queen of black card = Favorable outcomes/Total outcomes
= 2/52 = 1/26P(BQ) = 1/26
Question 2: What is the probability of getting a queen of spades?
Solution:
Total number of cards are 52 and number of queen of spades in 52 cards are 1.
So, total outcomes = 52
favorable outcomes = 1So, the probability of getting a queen of spades = Favorable outcomes/Total outcomes = 1/52
P(Q of S) = 1/52
Question 3: What is the probability of getting a heart of red cards?
Solution:
Total number of cards are 52 and number of heart jack card in 52 cards are 2.
So, total outcomes = 52
favorable outcomes = 2So, the probability of getting a heart of red card = Favorable outcomes/Total outcomes = 2/52 = 1/26
P(H of R) = 1/26
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Q:
Find the range of the data 2, 1, 2, 3, 5, 4, 7, 3, 5, 2, 4.
Answer & Explanation Answer: D) 6
Explanation:
View Answer Report Error Discuss
Playing cards probability problems based on a well-shuffled deck of 52 cards.
Basic concept on drawing a card:
In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. spades ♠ hearts ♥, diamonds ♦, clubs ♣.
Cards of Spades and clubs are black cards.
Cards of hearts and diamonds are red cards.
The card in each suit, are ace, king, queen, jack or knaves, 10, 9, 8, 7, 6, 5, 4, 3 and 2.
King, Queen and Jack (or Knaves) are face cards. So, there are 12 face cards in the deck of 52 playing cards.
Worked-out problems on Playing cards probability:
1. A card is drawn from a well shuffled pack of 52 cards. Find the probability of:
(i) ‘2’ of spades
(ii) a jack
(iii) a king of red colour
(iv) a card of diamond
(v) a king or a queen
(vi) a non-face card
(vii) a black face card
(viii) a black card
(ix) a non-ace
(x) non-face card of black colour
(xi) neither a spade nor a jack
(xii) neither a heart nor a red king
Solution:
In a playing card there are 52 cards.
Therefore the total number of possible outcomes = 52
(i) ‘2’ of spades:
Number of favourable outcomes i.e. ‘2’ of spades is 1 out of 52 cards.
Therefore, probability of getting ‘2’ of spade
Number of favorable outcomesP(A) = Total number of possible outcome = 1/52
(ii) a jack
Number of favourable outcomes i.e. ‘a jack’ is 4 out of 52 cards.
Therefore, probability of getting ‘a jack’
Number of favorable outcomesP(B) = Total number of possible outcome = 4/52 = 1/13
(iii) a king of red colour
Number of favourable outcomes i.e. ‘a king of red colour’ is 2 out of 52 cards.
Therefore, probability of getting ‘a king of red colour’
Number of favorable outcomesP(C) = Total number of possible outcome = 2/52 = 1/26
(iv) a card of diamond
Number of favourable outcomes i.e. ‘a card of diamond’ is 13 out of 52 cards.
Therefore, probability of getting ‘a card of diamond’
Number of favorable outcomesP(D) = Total number of possible outcome = 13/52 = 1/4
(v) a king or a queen
Total number of king is 4 out of 52 cards.
Total number of queen is 4 out of 52 cards
Number of favourable outcomes i.e. ‘a king or a queen’ is 4 + 4 = 8 out of 52 cards.
Therefore, probability of getting ‘a king or a queen’
Number of favorable outcomesP(E) = Total number of possible outcome = 8/52 = 2/13
(vi) a non-face card
Total number of face card out of 52 cards = 3 times 4 = 12
Total number of non-face card out of 52 cards = 52 - 12 = 40
Therefore, probability of getting ‘a non-face card’
Number of favorable outcomesP(F) = Total number of possible outcome = 40/52 = 10/13
(vii) a black face card:
Cards of Spades and Clubs are black cards.
Number of face card in spades (king, queen and jack or knaves) = 3
Number of face card in clubs (king, queen and jack or knaves) = 3
Therefore, total number of black face card out of 52 cards = 3 + 3 = 6
Therefore, probability of getting ‘a black face card’
Number of favorable outcomesP(G) = Total number of possible outcome = 6/52 = 3/26
(viii) a black card:
Cards of spades and clubs are black cards.
Number of spades = 13
Number of clubs = 13
Therefore, total number of black card out of 52 cards = 13 + 13 = 26
Therefore, probability of getting ‘a black card’
P(H) = Total number of possible outcome = 26/52 = 1/2
(ix) a non-ace:
Number of ace cards in each of four suits namely spades, hearts, diamonds and clubs = 1
Therefore, total number of ace cards out of 52 cards = 4
Thus, total number of non-ace cards out of 52 cards = 52 - 4
= 48
Therefore, probability of getting ‘a non-ace’
Number of favorable outcomesP(I) = Total number of possible outcome = 48/52 = 12/13
(x) non-face card of black colour:
Cards of spades and clubs are black cards.
Number of spades = 13
Number of clubs = 13
Therefore, total number of black card out of 52 cards = 13 + 13 = 26
Number of face cards in each suits namely spades and clubs = 3 + 3 = 6
Therefore, total number of non-face card of black colour out of 52 cards = 26 - 6 = 20
Therefore, probability of getting ‘non-face card of black colour’
Number of favorable outcomesP(J) = Total number of possible outcome = 20/52 = 5/13
(xi) neither a spade nor a jack
Number of spades = 13
Total number of non-spades out of 52 cards = 52 - 13 = 39
Number of jack out of 52 cards = 4
Number of jack in each of three suits namely hearts, diamonds and clubs = 3
[Since, 1 jack is already included in the 13 spades so, here we will take number of jacks is 3]
Neither a spade nor a jack = 39 - 3 = 36
Therefore, probability of getting ‘neither a spade nor a jack’
Number of favorable outcomesP(K) = Total number of possible outcome = 36/52 = 9/13
(xii) neither a heart nor a red king
Number of hearts = 13
Total number of non-hearts out of 52 cards = 52 - 13 = 39
Therefore, spades, clubs and diamonds are the 39 cards.
Cards of hearts and diamonds are red cards.
Number of red kings in red cards = 2
Therefore, neither a heart nor a red king = 39 - 1 = 38
[Since, 1 red king is already included in the 13 hearts so, here we will take number of red kings is 1]
Therefore, probability of getting ‘neither a heart nor a red king’
Number of favorable outcomesP(L) = Total number of possible outcome = 38/52 = 19/26
2. A card is drawn at random from a well-shuffled pack of cards numbered 1 to 20. Find the probability of
(i) getting a number less than 7
(ii) getting a number divisible by 3.
Solution:
(i) Total number of possible outcomes = 20 ( since there are cards numbered 1, 2, 3, ..., 20).
Number of favourable outcomes for the event E
= number of cards showing less than 7 = 6 (namely 1, 2, 3, 4, 5, 6).
So, P(E) = \(\frac{\textrm{Number of Favourable Outcomes for the Event E}}{\textrm{Total Number of Possible Outcomes}}\)
= \(\frac{6}{20}\)
= \(\frac{3}{10}\).
(ii) Total number of possible outcomes = 20.
Number of favourable outcomes for the event F
= number of cards showing a number divisible by 3 = 6 (namely 3, 6, 9, 12, 15, 18).
So, P(F) = \(\frac{\textrm{Number of Favourable Outcomes for the Event F}}{\textrm{Total Number of Possible Outcomes}}\)
= \(\frac{6}{20}\)
= \(\frac{3}{10}\).
3. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is
(i) a king
(ii) neither a queen nor a jack.
Solution:
Total number of possible outcomes = 52 (As there are 52 different cards).
(i) Number of favourable outcomes for the event E = number of kings in the pack = 4.
So, by definition, P(E) = \(\frac{4}{52}\)
= \(\frac{1}{13}\).
(ii) Number of favourable outcomes for the event F
= number of cards which are neither a queen nor a jack
= 52 - 4 - 4, [Since there are 4 queens and 4 jacks].
= 44
Therefore, by definition, P(F) = \(\frac{44}{52}\)
= \(\frac{11}{13}\).
These are the basic problems on probability with playing cards.
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