15 aptitude questions on probability, ranging from easy to moderate difficulty, along with detailed answers suitable for various competitive exams:
Question 1:
A bag contains 5 red balls and 3 blue balls. If one ball is drawn at random, what is the probability that the ball drawn is blue?
Answer:
Total number of balls: 5 (red) + 3 (blue) = 8
Number of blue balls: 3
Probability of drawing a blue ball: (Number of blue balls) / (Total number of balls) = 3/8
Question 2:
What is the probability of getting a number less than 5 when a fair six-sided die is rolled?
Answer:
Total possible outcomes: {1, 2, 3, 4, 5, 6} (6 outcomes)
Favorable outcomes (numbers less than 5): {1, 2, 3, 4} (4 outcomes)
Probability: (Favorable outcomes) / (Total possible outcomes) = 4/6=2/3
Question 3:
A coin is tossed twice. What is the probability of getting at least one head?
Answer:
Possible outcomes: {HH, HT, TH, TT} (4 outcomes)
Favorable outcomes (at least one head): {HH, HT, TH} (3 outcomes)
Probability: (Favorable outcomes) / (Total possible outcomes) = 3/4
Alternative approach: Probability of no heads (both tails) = 1/2x1/2=1/4. Probability of at least one head = 1 - (Probability of no heads) = 1−1/4=3/4.
Question 4:
From a well-shuffled deck of 52 cards, one card is drawn at random. What is the probability that the card is a king?
Answer:
Total number of cards: 52
Number of kings in a deck: 4 (one in each suit: hearts, diamonds, clubs, spades)
Probability: (Number of kings) / (Total number of cards) = 4/52=1/13
Question 5:
A box contains 7 green marbles and 5 yellow marbles. If two marbles are drawn at random without replacement, what is the probability that both marbles are green?
Answer:
Probability of the first marble being green: 7/12 (7 green out of 12 total)
After drawing one green marble, there are 6 green marbles left and 11 total marbles.
Probability of the second marble being green (given the first was green):6/11
Probability of both marbles being green: 7/12×6/11=42/132=21/66=7/22
Question 6:
In a class of 40 students, 60% are girls. If one student is selected at random, what is the probability that the selected student is a boy?
Answer:
Number of girls: 60% of 40 = 0.60×40=24
Number of boys: Total students - Number of girls = 40−24=16
Probability of selecting a boy: (Number of boys) / (Total number of students) = 16/40=2/5
Question 7:
Two dice are rolled simultaneously. What is the probability that the sum of the numbers on the two dice is 7?
Answer:
Total possible outcomes: 6×6=36 (each die has 6 faces)
Favorable outcomes (sum is 7): {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} (6 outcomes)
Probability: (Favorable outcomes) / (Total possible outcomes) = 6/36=1/6
Question 8:
A committee of 3 people is to be formed from a group of 4 men and 3 women. What is the probability that the committee will consist of 2 men and 1 woman?
Answer:
Total number of ways to form a committee of 3 from 7 people: (37)=3!(7−3)!7!=3×2×17×6×5=35
Number of ways to choose 2 men from 4: (24)=2!(4−2)!4!=2×14×3=6
Number of ways to choose 1 woman from 3: (13)=1!(3−1)!3!=13=3
Number of ways to form a committee of 2 men and 1 woman: (24)×(13)=6×3=18
Probability: (Favorable outcomes) / (Total possible outcomes) = 18/35
Question 9:
A letter is chosen at random from the word "PROBABILITY". What is the probability that the chosen letter is a vowel?
Answer:
Total number of letters in "PROBABILITY": 11
Number of vowels (A, E, I, O, U) in "PROBABILITY": O, A, I (3 vowels)
Probability: (Number of vowels) / (Total number of letters) = 3/11
Question 10:
Events A and B are such that P(A)=0.4, P(B)=0.3, and P(A∩B)=0.2. Find P(A∪B).
Answer:
We use the formula for the probability of the union of two events: P(A∪B)=P(A)+P(B)−P(A∩B)
Substituting the given values: P(A∪B)=0.4+0.3−0.2=0.7−0.2=0.5
Question 11:
A bag contains 8 white balls and some black balls. If the probability of drawing a black ball is twice that of drawing a white ball, find the number of black balls.
Answer:
Let the number of black balls be x.
Total number of balls: 8+x
Probability of drawing a white ball: 8+x8
Probability of drawing a black ball: 8+xx
According to the problem, P(black)=2×P(white)
8+xx=2×8+x8
Multiplying both sides by (8+x): x=16
Therefore, the number of black balls is 16.
Question 12:
What is the probability that a leap year, selected at random, will contain 53 Sundays?
Answer:
A leap year has 366 days, which is 52 weeks and 2 extra days.
These 2 extra days can be any of the following pairs: (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday).
Out of these 7 equally likely pairs, 2 of them contain a Sunday.
Therefore, the probability that a leap year will contain 53 Sundays is 72.
Question 13:
A card is drawn from a standard deck of 52 cards. What is the probability that the card is either a heart or a face card (King, Queen, Jack)?
Answer:
Number of hearts: 13
Number of face cards: 12 (4 Jacks, 4 Queens, 4 Kings)
Number of cards that are both a heart and a face card: 3 (Jack of hearts, Queen of hearts, King of hearts)
We use the formula P(A∪B)=P(A)+P(B)−P(A∩B)
P(heart)=13/52
P(face card)=12/52
P(heart and face card)=3/52
P(heart or face card)=13/52+12/52−3/52=13/52+12−3=22/52=11/26
Question 14:
A and B are two independent events such that P(A)=0.5 and P(B)=0.4. Find the probability that neither A nor B occurs.
Answer:
If A and B are independent, then P(A∩B)=P(A)×P(B)=0.5×0.4=0.2.
The probability that neither A nor B occurs is P(A∩B), which is equal to P(A∪B) by De Morgan's Law.
P(A∪B)=P(A)+P(B)−P(A∩B)=0.5+0.4−0.2=0.7.
P(A∪B)=1−P(A∪B)=1−0.7=0.3.
Alternative approach:
P(A)=1−P(A)=1−0.5=0.5
P(B)=1−P(B)=1−0.4=0.6
Since A and B are independent, A and B are also independent.
P(A∩B)=P(A)×P(B)=0.5×0.6=0.3.
Question 15:
A bag contains 4 white balls, 5 black balls, and 6 red balls. If two balls are drawn at random, what is the probability that one is white and the other is black?
Answer:
Total number of balls: 4+5+6=15
Number of ways to choose 2 balls from 15=105
Number of ways to choose one white ball from 4=4
Number of ways to choose one black ball from 5)=5
Number of ways to choose one white and one black ball: )=4×5=20
Probability: (Favorable outcomes) / (Total possible outcomes) = 20/105=4/21
