In a pack of cards, Ace, King, Queen, and Jack are called honours. There are 16 honours in a pack . An honour can be drawn in 16C1 ways (i.e., n(E)=16). One card can be drawn from a pack in 52C1 ways. Required probability is 16/52=4/13. The application or uses of probability can be seen in quantitative aptitude as well as in daily life. It is needful to learn the basic concept of probability. We will cover the basics as well as the hard level problems for all levels of students for all competitive exams especially SBI PO, SBI CLERK, IBPS PO, IBPS CLERK, RRB PO, NICL AO, LIC AAO, SNAP, MAT, SSC CGL etc.
Definition: Probability means the possibility
or chances of an event occurring or happening. For example, when a coin is tossed, then we will get ahead or tail. It is a state of probability. In an event, the happening probability is equal to the ratio of favourable outcomes to the total number of possible outcomes. It represents as, Number of favourable outcomes = ____________________________________ Total number of
possible outcomes Sample Space:- It is a set of all possible outcomes of an experiment. It is denoted by S. For example, the Sample space of a die, S = [ 1, 2, 3 , 4, 5, 6] The Sample space of a coin, S= [ Head, Tail] Types of questions asked in the competitive exam: 1) Based on Coins 2) Based on Dice 3) Based on
playing Cards 4) Based on Marbles or balls 5) Miscellaneous Important Questions: 1. Question A coin is thrown two times .what is the probability that at least one tail is obtained? A) 3/4 B) 1/4 C) 1/3 D) 2/3 E) None of these Answer :- A Sol: Sample space = [TT, TH, HT,HH] Total number of ways =
2 × 2 = 4. Favourite Cases = 3 P (A) = 3/4 Tricks:- P (of getting at least one tail) = 1 – P (no head)⇒ 1 – 1/4 = 3/4 2. Question What is the probability of getting a numbered card when drawn from the pack of 52 cards? A) 1/13 B) 1/9 C) 9/13 D) 11/13 E) None of these Answer :- C Sol: Total Cards = 52. Numbered Cards = 9
(2,3,4,5,6,7,8,9,10) in each suit Numbered cards in four suit = 4 ×9 = 36 P (E) = 36/52 = 9/13 3.Question There are 7 purple clips and 5 brown clips. Two clips are selected one by one without replacement. Find the probability that the first is brown and the second is purple. A) 1/35 B) 35/132 C) 1/132 D) 35/144 E) None of these Answer :- B Sol: P (B) × P (P) = (5/12) x (7/11) =
35/132 4.Question Find the probability of getting a sum of 8 when two dice are thrown? A) 1/8 B) 1/5 C) 1/4 D) 5/36 E) 1/3 Answer 😀 Sol: Total number of ways = 6 × 6 = 36 ways. Favorable cases = (2 , 6) (6, 2) (3, 5) (5, 3) (4, 4) — 5 ways. P (A) = 5/36 = 5/36 5.Question Find the probability of an honour card when a card is drawn at random
from the pack of 52 cards. A) 4/13 B) 1/3 C) 5/12 D) 7/52 E) None of these Answer :-A Sol: Honor cards = 4 (A, J, Q, K) in each suit Honor cards in 4 suit = 4 × 4 = 16 P (honor card) = 16/52 = 4/13 6. Question What is the probability of a face card when a card is drawn at random from the pack of 52 cards? A) 1/13 B) 2/13 C) 3/13 D) 4/13 E) 5/13 Answer
:-C Solution: face cards = 3 (J,Q,K) in each suit Face cards in 4 suits = 3 × 4 = 12 Cards. P (face Card) = 12/52 = 3/13 7.Question If two dice are rolled together then find the probability as getting at least one ‘3’? A) 11/36 B) 1/12 C) 1/36 D) 13/25 E) 13/36 Answer :- A Sol: Total number of ways = 6 × 6 = 36. Probability of getting number
‘3′ at least one time = 1 – (Probability of getting no number 4) = 1 – (5/6) x (5/6) = 1 – 25/36 = 11/36 8. Question If a single six-sided die is rolled then find the probability of getting either 3 or 4. A) 1/2 B) 1/3 C) 1/4 D) 2/3 E) 1/6 Answer:- B Solution:- Total outcomes = 6 Probability of getting a single number when rolled a die = 1/6 So, P(3) = 1/6 and P(4) = 1/6 Thus, the
probability of getting either 3 or 4 = P(3)+P(4) = 1/6 + 1/6 = 1/3 9. Question A container contains 1 red, 3 black, 2 pink and 4 violet gems. If a single gem is chosen at random from the container, then find the probability that it is violet or black? A) 1/10 B) 3/10 C) 7/10 D) 9/10 E) None of these Answer :-C Sol :- Total gems =( 1 + 3 + 2 + 4 ) = 10 probability of getting a violet gem =
4/10 Probability of getting a black gem = 3/10 Now, P ( Violet or Black) = P(violet) + P(Black) = 4/10 + 3/10 = 7/10 10.Question A jar contains 63 balls (
1,2,3,……., 63). Two balls are picked at random from the jar one after one and without any replacement. what is the probability that the sum of both balls drawn is even? A) 5/21 B) 3/23 C) 5/63 D) 19/63 E) None of these Answer :- A Sol. Total balls = 63 Total even balls = 31 ( 2 , 4 , 6,……., 62) Now the required probability =³¹C₂/63C₂
= (31!/2!29!)/(63!/2!61!) = (31 × 30/1× 2)/(63×62/1×2) = (31 × 30)/(63×62) = 30/63×2 = 5/21 11.Question There are 30 students in a class, 15 are boys and 15 are girls. In the final exam, 5 boys and 4 girls made an A grade. If a student is chosen at random from the class, what is the probability of choosing a girl or an ‘A grade student? A) 1/4 B)
3/10 C) 1/3 D) 2/3 E) None of these Answer:- D Sol: Here, the total number of boys = 15 and the total number of girls = 15 Also, girls getting A grade = 4 and boys getting an A grade = 5 Probability of choosing a girl = 15/30 Probability of choosing A grade student= 9/30 Now, an A-grade student chosen can be a girl. So the probability of choosing it = 4/30 Required probability of choosing a girl or an A
grade student = 15/30 + 9/30 – 4/30 = 1/2 + 3/10 – 2/15 = 2/3 12. Question What is the probability when a card is drawn at random from a deck of 52 cards is either an ace or a club? A) 2/13 B) 3/13 C) 4/13 D) 5/23 E) None of these Answer:-
C Sol: There are 4 aces in a pack, 13 club cards and 1 ace of club card. Now, the probability of getting an ace = 4/52 Probability of getting a club = 13/52 Probability of getting an ace of club = 1/52 Required probability of getting an ace or a club = 4/52 + 13/52 – 1/52 = 16/52 = 4/13 13. Question One card is drawn from a deck of 52 cards well shuffling. Calculate the probability that the card will
not be a king. A) 12/13 B) 3/13 C) 7/13 D) 5/23 E) None of these Answer:- A Solution: Well-shuffling ensures equally likely outcomes. Total king of a deck = 4 The number of favourable outcomes F= 52 – 4 = 48 The number of possible outcomes = 52 Therefore, the required probability = 48/52 = 12/13 14.Question If P(A) = 7/13, P(B) = 9/13 and P(A∩B) = 4/13, find the
value of P(A|B). A) 1/9 B) 2/9 C) 3/9 D) 4/9 E) None of these Answer :- D Solution: P(A|B) = P(A∩B)/P(B) = (4/13)/(9/13) = 4/9. 15. Question A one rupee coin and a two rupee coin are tossed once, then calculate a sample space. A) [ HH, HT, TH, TT] B) [ HH, TT] C) [ TH, HT] D) [HH, TH, TT] E) None of these Answer:- A Solution:
The outcomes are either Head (H) or tail(T). Now,heads on both coins = (H,H) = HH Tails on both coins = ( T, T) = TT Probability of head on one rupee coin and Tail on the two rupee coins = (H, T) = HT And Tail on one rupee coin and Head on the two rupee coin = (T, H) = TH Thus, the sample space ,S = [HH, HT, TH, TT] 16. Question There are 20 tickets numbered 1 to 20. These tickets are mixed up and then a ticket is drawn at random.
Find the probability that the ticket drawn has a number which is a multiple of 4 or 5? A) 1/4 B) 2/13 C) 8/15 D) 9/20 E) None of these Answer: D Solution: Here, S = {1, 2, 3, 4, …., 19, 20} = 20 Let E = event of getting a multiple of 4 or 5 = {4, 8 , 12, 16, 20, 5, 10, 15, 20} = 9 Required probability = favourable outcomes/total outcomes = 9/20 Direction ( 17 – 19):- In a
school the total number of students is 300, 95 students like chicken only, 120 students like fish only, 80 students like mutton only and 5 students do not like anything above. If randomly one student is chosen, find the probability that 17) The student likes mutton. 18 ) he likes either chicken or mutton 19 ) he likes neither fish nor mutton. Solution( 17-19):- The total number of favourable outcomes = 300 (Since there are 300 students
altogether).
The number of times a chicken liker is chosen = 95 (Since 95 students like chicken). The number of times a fish liker is chosen = 120. The number of times a mutton liker is chosen = 80. The number of times a student is chosen who likes none of these = 5. 17. Question Find the probability that the student like mutton? A) 3/10 B) 4/15 C) 1/10 D) 1/15 E) None of these Answer:- B Solution:-
Therefore, the probability of getting a student who likes mutton = 80/300 = 4/15 18. Question What is the probability that the student likes either chicken or mutton? A) 7/12 B) 5/12 C) 3/4 D) 1/12 E) None of these Answer:- A Solution:- The probability of getting a student who likes either chicken or mutton = (95+80)/300 = 175/300 = 7/12 19.
Question Find the probability that the student likes neither fish nor mutton. A) 1/2 B) 1/5 C) 1/3 D) 1/4 E) 1/6 Answer:- C Solution:- The probability of getting a student who likes neither fish nor mutton = (300–120−80)/300 = 100/300 = 1/3 Direction ( 20-22):- A box contains 90 number plates numbered 1 to 90. If one number plate is drawn at random from the box then find out the
probability that 20) The number is a two-digit number 21) The number is a perfect square 22) The number is a multiply of 5 20. Question Find the probability that the number is a two-digit number. A) 1/9 B) 1/10 C) 9/10 D) 7/10 E) None of these Answer:-C Solution : Total possible outcomes = 90 (Since the number plates are numbered from 1 to 90). Number of favourable outcomes = 90 –
9 = 81 ( here, except 1 to 9, other numbers are two-digit number.) Thus required probability = Number of Favourable Outcomes /Total Number of Possible Outcomes = 81/90 = 9/10. 21. Question What is the probability that the number is a perfect square? A) 1/9 B) 1/10 C) 9/10 D) 1/7 E) None of these Answer:- B Solution:- Total possible outcomes = 90. Number of favourable outcomes =
9 [here 1, 4, 9, 16, 25, 36, 49, 64 and 81 are the perfect squares] Thus the required probability = 9/90 =1/10 22.Question Find the probability that the number is a multiple of 5. A) 1/5 B) 1/6 C) 1/10 D) 1/8 E) 9/10 Answer:-
A Solution:- Total possible outcomes = 90. Number of favourable outcomes = 18 (here, 5 × 1, 5 × 2, 5 × 3, …., 5 × 18 are multiple of 5). Thus, the required probability= 18/90 =1/5
How many honour cards are there in a pack of cards?
Therefore total number of honour cards =4×4=16.
What is the chance of selecting two queens one by one from the pack of 52 cards without replacement?
To find the P(QQQ), we find the probability of drawing the first queen which is 4/52. The probability of drawing the second queen is also 4/52 and the third is 4/52.
What is the probability of drawing 2 spades from a well shuffled pack of 52 cards?
Hence the probability of getting '2' of spades is 1/52.
What is the probability of getting an honor card when a card is drawn from the pack of 52 cards?
Detailed Solution. ∴ The probability to get honor card is 4/13.
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