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How To Solve Probability Questions Quickly
How to Solve Probability Questions Quickly
Here , In this Page you learn how to Solve Probability Questions Quickly. Go through this page to get sample Probability questions to get the clear understanding of method to solve the questions based on probability.
Therefore, probability of the occurrence of event is the number between 0 and 1..
How to Solve Quickly Probability Questions
 You can solve many simple probability problems just by knowing two simple rules:
 The probability of any sample point can range from 0 to 1.
 The sum of probabilities of all sample points in a sample space is equal to 1.
 The probability of event A is denoted by P(A).
Formula of Probability and their related concepts:

Probability of an event (A): P(A) = \frac{Number.of.favorable.outcomes}{Number.of.possible.outcomes}
This formula calculates the probability of an event A occurring. You count the number of outcomes that satisfy event A and divide it by the total number of possible outcomes.

Addition rule: P(A or B) = P(A) + P(B) – P(A and B)
This formula calculates the probability of event A or event B occurring. You sum the probabilities of each event individually, but subtract the probability of both events occurring simultaneously to avoid doublecounting.

Multiplication rule (for independent events): P(A and B) = P(A) * P(B)
This formula calculates the probability of event A and event B occurring simultaneously, assuming that the events are independent. You multiply the probabilities of each event individually.

Multiplication rule (for dependent events): P(A and B) = P(A) * P(B  A)
This formula calculates the probability of event A and event B occurring simultaneously, assuming that the events are dependent. P(BA) denotes the probability of event B occurring given that event A has already occurred.

Conditional probability: P(A  B) = \frac{P(A.and.B)}{P(B)}
This formula calculates the probability of event A occurring given that event B has already occurred. P(A and B) denotes the probability of both events A and B occurring simultaneously, and P(B) represents the probability of event B occurring.

Complementary probability: P(A’) = 1 – P(A)
This formula calculates the probability of the complement of event A, denoted as A’. It subtracts the probability of event A occurring from 1, yielding the probability of event A not occurring.

Bayes’ theorem: P(AB) = P(BA)\times\frac{P(A)}{P(B)}
Bayes’ theorem allows you to calculate the probability of event A occurring given event B, by utilizing conditional probabilities. P(AB) represents the probability of event A given event B, P(BA) is the probability of event B given event A, P(A) is the prior probability of event A, and P(B) is the prior probability of event B.
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Types 1 How to Solve Probability Questions Quickly of Random ticket or ball drawn
Question 1. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
Options
(a) \frac{1}{2}
(b) \frac{2}{5}
(c) \frac{8}{15}
(d) \frac{9}{20}
Solution: Here, S = {1, 2, 3, 4, …., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = \frac{n(E)}{n(S)} = \frac{9}{20}
Correct Options (D).
Question 2 A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Options
(a) \frac{10}{21}
(b) \frac{11}{21}
(c) \frac{2}{27}
(d) \frac{5}{7}
Solution: Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7.
= ^{7}C_{2}
= \frac{7 × 6}{2 × 1}
= 21.
Let E = Event of drawing 2 balls, none of which is blue.
n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls.
= ^{5}C_{2}
= \frac{5 × 4}{2 × 1}
= 10 .
P(E) = \frac{n(E)}{n(S)} = \frac{10}{21} .
Correct Options (A)
Questions 3 A bag contains 1100 tickets numbered 1, 2, 3, … 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it?
(a) \frac{291}{1100}
(b) \frac{292}{1100}
(c) \frac{290}{1100}
(d) \frac{301}{1100}
Solution: Ticket has a maximum of 4 digits on it as Thousands, Hundreds Tens ,Units or TH H T U.
Number of Tickets with 2 in TH place = 0.
Number of Tickets with 2 in H place = From 200 upto 299 = 100.
Number of Tickets with 2 in T place but not in H place = 20 to 29 in T and U places and 00 to 10 except 02 in TH and H places = 10×10 = 100.
Number of Tickets with 2 ONLY in U place but not in TH H or T place
= H or T place both have (0 to 9 excluding 2) + (TH=1 & H=0 & U
= 2 & (T= 0 to 9 excluding 2) )
= (9×9) + 9 = 90.
Total Tickets with at least one 2 = 290
Probability = \frac{290}{1100} is answer.
Correct Options (c)
Type 2 How to Solve Probability Questions Quickly of boys and girls
Question 1. In a class there are 60% of girls of which 25% poor. What is the probability that a poor girl is selected is leader?
Options
(a) 20%
(b) 15%
(c) 10%
(d) 25%
Solutions: Assume total students in the class = 100
Then Girls = 60% (100) = 60
Poor girls = 25% (60) = 15
So probability that a poor girls is selected leader = \frac{Poor girls}{ Total students} = \frac{15}{100} = 15%
Correct Options (b)
Questions 2. What is the probability that the total of two dice will be greater than 9, given that the first die is a 5?
Options
(a) \frac{1}{3}
(b) \frac{1}{6}
(c) \frac{1}{9}
(d) None of these
Solution : Let A= first die is 5
Let B = total of two dice is greater than 9
P(A) = Possible outcomes for A and B: (5, 5), (5, 6)
P(A and B) = \frac{2}{36} = \frac{1}{18}
P(BA) = \frac{P(A and B)}{P(A)} = \frac{1}{18} ÷ \frac{1}{6} = \frac{1}{3}.
Correct Options (A)
Questions 3. If six cards are selected at random (without replacement) from a standard deck of 52 cards, what is the probability there will be no pairs? (two cards of the same denomination)
Solution: Let E_{i} be the event that the first i cards have no pair among them. Then we want to compute P(E_{6}).
Which is actually the same as P(E_{1} ∩ E_{2} ∩ · · · ∩ E_{6}), since E_{6} ⊂ E_{5} ⊂ · · · ⊂ E_{1}, implying that E_{1} ∩ E_{2} ∩ · · · ∩ E_{6} = E_{6}.
We get P(E_{1} ∩ E_{2} ∩ · · · ∩ E_{6}) = P(E_{1})P(E_{2}E_{1})· · · = \frac{52}{52} \frac{48}{51} \frac{44}{50} \frac{40}{49} \frac{36}{48} \frac{32}{47}
Alternatively, one can solve the problem directly using counting techniques.
Define the sample space to be (equally likely)ordered sequences of 6 cards; then, S = 52 · 51 · 50 · · · 47, and the event E6 has 52 · 48 · 44 · · · 32 elements
Questions 4. A group of 5 friendsArchie, Betty, Jerry, Moose, and Veronicaarrived at the movie theater to see a movie. Because they arrived late, their only seating option consists of 3 middle seats in the front row, an aisle seat in the front row, and an adjoining seat in the third row. If Archie, Jerry, or Moose must sit in the aisle seat while Betty and Veronica refuse to sit next to each other, how many possible seating arrangements are there?
Options
(a) 32
(b) 36
(c) 48
(d) 72
(e) 120
Solution: Good = Total – Bad.
Total = arrangements with Archie, Jerry or Moose in the aisle seat.
Number of options for the aisle seat = 3. (Archie, Jughead, or Moose)
Number of ways to arrange the 4 other people = 4 x 3 x 2 x 1.
To combine these options, we multiply: 3 x 4 x 3 x 2 = 72.
Bad = arrangements with Archie, Jerry or Moose in the aisle seat BUT with Betty next to Veronica.
Number of options for the aisle seat = 3. (Archie, Jughead, Moose).
Number of options for the third row seat = 2. (Anyone but Betty and Veronica, since in a bad arrangement they sit next to each other.)
Number of options for the middle of the 3 remaining seats = 2. (Must be Betty or Veronica so that they sit next to each other).
Number of ways to arrange the 2 remaining people = 2 x 1.
To combine these options, we multiply: 3 x 2 x 2 x 2 = 24.
Good arrangements = 72 – 24 = 48.
Correct Option C.
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