Probability Theory

Introduction to Probability

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Probability undoubtedly is a vast topic with applications in various fields. With countless applications in research, sports, games, weather forecast, insurance, business analytics, etc, probability is one of the most important topics that we must be aware of.

The aim of this topic is to help you build your basics in Probability.  In the section introduction to Probability we have shared some important terminologies involved in Probability problems. The exercises contain solved question and answers. The key to solving a problem is knowing how to approach the problem. When solving exercises, we recommend you to try solving the problems on your own, then view the solutions. That way we believe you can strengthen your approach in solving problems on Probability.

Introduction to Probability

Experiment :

An experiment is a trial that gives well defined set of possible outcomes.

Outcome : 

It is the result of single trial of an experiment.

Event : 

Event is one or more outcomes of an experiment.

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Random Experiment :

An experiment whose outcome cannot be predicted in advance with certainty.

Example : Tossing a coin is a random experiment whose outcome is either heads or tails.

Sample Space(S) : 

Set of all possible outcomes of an experiment.

Example : Sample space of a rolling a dice – {1, 2, 3, 4, 5, 6}

Event : 

It is the subset of a sample space.

Probability : 

Probability is the measure of likelihood that an event will occur.

Probability of an event A is the number of ways event A can occur divided by the total number of possible outcomes.
i.e, Probability of event A = $\frac{\text{Number of ways event A can occur}}{\text{Total number of possible outcomes}} $

P(A) = $\frac{n(A)}{n(S)}$

where,
P(A) is the probability of event A
n(A) is number of ways event can occur,
n(S) is the total number of possible outcomes

About Playing Cards

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.

Problems

Practice

Probability - Exercise - Blog

19 questions

  1. What is the probability of getting a prime number when a dice is rolled?
    • $\frac{1}{6}$
    • $\frac{1}{2}$
    • $\frac{2}{3}$
    • $\frac{3}{5}$
    Show the answer

    Answer: $\frac{1}{2}$

    S : sample space.
    n(S) : Total number of possible outcomes when a dice is rolled = 6

    E : getting a prime number when a dice is rolled = {2,3,5}
    n(E) = 3

    P(E) = $\frac{n(E)}{n(S)}$ = $\frac{3}{6}$ = $\frac{1}{2}$

  2. Two dice are rolled, what is the probability that the sum of the numbers is either 3 or 5?
    • $\frac{1}{6}$
    • $\frac{1}{4}$
    • $\frac{1}{2}$
    • $\frac{2}{3}$
    Show the answer

    Answer: $\frac{1}{6}$

    Number of possible outcomes when a dice is rolled : 6
    S : Sample space representing possible outcomes when two dice are rolled
    n(S) : Total number of possible outcomes when two dice are rolled = 6x6 = 36

    E1: Sum of the numbers on the two dice is 3 = {(1,2), (2,1)}
    n(E1) = 2

    p(E1) = $\frac{n(E1)}{p(E1)}$ = $\frac{2}{36}$

    E2: Sum of the numbers on the two dice is 5 = {(1,4), (2,3), (3,2), (4,1)}
    n(E2) = 4

    p(E1) = $\frac{n(E2)}{p(E2)}$ = $\frac{4}{36}$

    E1 and E2 are mutually exclusive events. Hence, By Addition theorem of probability

    p(S) = p(E1) + p(E2) = $\frac{2}{36}$ + $\frac{4}{36}$ = $\frac{1}{6}$

  3. What is the probability of getting a sum that is a multiple of 2 or 3 when two dice are rolled together?
    • $\frac{1}{6}$
    • $\frac{1}{4}$
    • $\frac{1}{2}$
    • $\frac{2}{3}$
    Show the answer

    Answer: $\frac{2}{3}$

    S : Sample space representing possible outcomes when two dice are rolled
    n(S) : Total number of possible outcomes when two dice are rolled = 6x6 = 36

    E : Sum that is multiple of 2 or 3
    E1 : Sum that is multiple of 2
    n(E1) = 18
    E2 : Sum that is multiple of 3
    n(E2) = 12
    E3 : Sum that is multiple of 2 and 3
    n(E3) = 6

    p(E) = p(E1) + p(E2) - p(E3) [Non-Mutually Exclusive events]

    p(E1) = $\frac{n(E1)}{n(S)}$ = $\frac{18}{36}$ = $\frac{1}{2}$

    p(E2) = $\frac{n(E2)}{n(S)}$ = $\frac{12}{36}$ = $\frac{1}{3}$

    p(E3) = $\frac{n(E3)}{n(S)}$ = $\frac{6}{36}$ = $\frac{1}{6}$

    p(E) = $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{6}$ = $\frac{2}{3}$

  4. Two dice are rolled together. What is the probability of getting two numbers whose product is odd.
    • $\frac{1}{5}$
    • $\frac{3}{5}$
    • $\frac{1}{4}$
    • $\frac{3}{4}$
    Show the answer

    Answer: $\frac{1}{4}$

    S : Sample space representing possible outcomes when two dice are rolled
    n(S) : Total number of possible outcomes when two dice are rolled = 6x6 = 36

    E : getting two numbers whose product is odd = {(1,1), (1,3), (1,5), (3,1), (3, 3), (3,5), (5,1), (5,3), (5,5)}
    n(E) = 9

    p(E) = $\frac{n(E)}{n(S)}$ = $\frac{9}{36}$ = $\frac{1}{4}$

  5. A dice is rolled 3 times. What is the probability of getting same face in all 3 throws of the dice?
    • $\frac{1}{6}$
    • $\frac{1}{36}$
    • $\frac{1}{216}$
    • $\frac{1}{9}$
    Show the answer

    Answer: $\frac{1}{36}$

    S : Sample space representing possible outcomes when two dice are rolled
    n(S) : Total number of possible outcomes when dice is rolled 3 times = $6^3$

    E : Getting same face in all 3 throws of the dice
    n(E) = $^6C_1$

    p(E) = $\frac{^6C_1}{6^3}$ = $\frac{1}{6^2}$ = $\frac{1}{36}$

  6. A bag contains 5 white and 3 green balls. A ball is drawn at random. What is the probability that the ball drawn is white.
    • $\frac{3}{4}$
    • $\frac{2}{3}$
    • $\frac{5}{8}$
    • $\frac{1}{2}$
    Show the answer

    Answer: $\frac{5}{8}$

    S : Sample space represents drawing of 1 ball from 8 balls (5 white + 3 green)
    n(S) : Total number of ways of drawing 1 ball from 8 balls = $^8C_1$ = 8

    E : An event representing drawing of white ball
    n(E) : Number of ways of drawing a white ball from 5 white balls = $^5C_1$ = 5

    p(E) = $\frac{n(E)}{n(S)}$ = $\frac{5}{8}$

  7. A bag contains 5 blue balls, 4 green balls. 2 balls are drawn at random. What is the probability that the balls drawn are of same colour?
    • $\frac{1}{9}$
    • $\frac{2}{9}$
    • $\frac{3}{9}$
    • $\frac{4}{9}$
    Show the answer

    Answer: $\frac{4}{9}$

    S : Sample space represents drawing of 2 balls from 9 balls (5 blue + 4 green)
    n(S) : Total number of ways of drawing 2 balls from 9 balls = $^9C_2$ = 36

    E : Event representing drawing 2 balls of same colour (blue or green)
    n(E) : Number of ways of drawing 2 balls of blue colour + Number of ways of drawing 2 balls of green colour

    E1 : An event representing drawing 2 balls of blue colour
    n(E1) : Number of ways of drawing 2 blue ball from 5 blue balls = $^5C_2$ = 10

    E2 : An event representing drawing 2 balls of green colour
    n(E2) : Number of ways of drawing 2 green balls from 4 green balls = $^4C_2$ = 6

    n(E) = n(E1) + n(E2) = 10 + 6 = 16

    p(E) = $\frac{n(E)}{n(S)}$ = $\frac{16}{36}$ = $\frac{4}{9}$

  8. A bag contains 4 black, 6 blue and 5 yellow balls. Two balls are drawn at random. What is the probability that the balls drawn are neither black nor yellow.
    • $\frac{1}{7}$
    • $\frac{4}{15}$
    • $\frac{1}{3}$
    • $\frac{2}{3}$
    Show the answer

    Answer: $\frac{1}{7}$

    S : Sample space represents drawing of 2 balls from 15 balls (4 black + 6 blue + 5 yellow)
    n(S) : Total number of ways of drawing 2 balls from 15 balls = $^15C_2$ = 105

    E : Event representing drawing 2 balls of blue colour (neither black nor yellow)
    n(E) : Number of ways of drawing 2 blue coloured balls from 6 blue coloured balls = $^6C_2$ = 15

    p(E) = $\frac{n(E)}{n(S)}$ = $\frac{15}{105}$ = $\frac{1}{7}$

  9. 3 balls are randomly drawn from a bag containing 3 red, 4 blue and 5 green balls. What is the probability that the balls drawn are of different colours?
    • $\frac{2}{5}$
    • $\frac{5}{9}$
    • $\frac{3}{11}$
    • $\frac{5}{13}$
    Show the answer

    Answer: $\frac{3}{11}$

    S : Sample space represents drawing of 3 balls from 12 balls (3 red + 4 blue + 5 green)
    n(S) : Total number of ways of drawing 3 balls from 12 balls = $^12C_3$ = 220

    E : Event representing drawing 3 different coloured balls
    n(E) : Number of ways of drawing 1 red ball from 3 red balls, 1 blue ball from 4 blue balls and 1 green ball from 5 green bals = $^3C_1$ x $^4C_1$ x$^5C_1$ = $\frac{60}{220}$ = $\frac{3}{11}$

  10. What is the probability of drawing a card of diamond from a pack of 52 cards?
    • $\frac{1}{13}$
    • $\frac{1}{4}$
    • $\frac{1}{2}$
    • $\frac{3}{4}$
    Show the answer

    Answer: $\frac{1}{4}$

    S : Drawing of 1 card from 52 cards
    n(S) : Number of ways of drawing 1 card from 52 cards = 52

    Number of diamond cards = 13

    E : Drawing a diamond card
    n(E) : 13

    p(E) = $\frac{n(E)}{n(S)}$ = $\frac{13}{52}$ = $\frac{1}{4}$

  11. From a pack of 52 cards, a card is drawn at random. What is the probability that the card drawn is either king or queen?
    • $\frac{1}{13}$
    • $\frac{2}{13}$
    • $\frac{3}{13}$
    • $\frac{4}{13}$
    Show the answer

    Answer: $\frac{2}{13}$

    S : All the 52 playing cards
    n(S) : Total number of possible outcomes = 52

    K : Event of drawing a King
    n(K) : Total number of King = 4

    Q : Event of drawing a queen
    n(Q) : Total number of queen = 4

    E : Drawing a king or queen
    n(E) = n(K) + n(Q) = 4 + 4 = 8

    p(E) = $\frac{n(E)}{n(S)}$ = $\frac{8}{52}$ = $\frac{2}{13}$

  12. Two cards are randomly drawn from a pack of 52 cards. What is the probability that one of the card drawn is spade and the other one is club.
    • $\frac{2}{13}$
    • $\frac{10}{99}$
    • $\frac{13}{102}$
    • $\frac{10}{101}$
    Show the answer

    Answer: $\frac{13}{102}$

    S : Drawing of 2 cards from 52 playing cards
    n(S) : Total number of ways of drawing 2 cards from 52 cards = $^{52}C_2$ = 1326

    E : Drawing of 1 spade and 1 club
    Total number of spades = 13
    Total number of clubs = 13
    n(E) : Number of ways of drawing 1 spade and 1 diamond = $^{13}C_1$ x $^{13}C_1$ = 169

    p(E) = $\frac{n(E)}{n(S)}$ = $\frac{169}{1326}$ = $\frac{13}{102}$

  13. A card is drawn from a pack of 52 cards. What is the probability of drawing a black face card.(King, Queen or Jack)
    • $\frac{1}{9}$
    • $\frac{3}{26}$
    • $\frac{2}{13}$
    • $\frac{4}{13}$
    Show the answer

    Answer: $\frac{3}{26}$

    S : Drawing of 1 card from 52 cards
    n(S) : Number of ways of drawing 1 card from 52 cards = 52

    Cards of spades and clubs are black cards
    Number of face cards in spades = 3
    Number of face cards in clubs = 3
    Total number of black face cards = 6

    E : Drawing a black face card
    n(E) : Number of ways of drawing black face card = 6

    p(E) = $\frac{n(E)}{n(S)}$ = $\frac{6}{52}$ = $\frac{3}{26}$

  14. What is the probability of drawing a 2 of heart or diamond?
    • $\frac{1}{26}$
    • $\frac{1}{13}$
    • $\frac{3}{13}$
    • $\frac{3}{26}$
    Show the answer

    Answer: $\frac{1}{26}$

    S : Drawing of 1 card from 52 cards
    n(S) : Number of ways of drawing 1 card from 52 cards = 52

    Number of heart cards of number 2 = 1
    Number of diamond cards of number 2 = 1

    E : Drawing a 2 of heart or diamond
    n(E) : 1 + 1 = 2

    p(E) = $\frac{n(E)}{n(S)}$ = $\frac{2}{52}$ = $\frac{1}{26}$

  15. What is the probability that letter chosen from a set of english alphabets is a vowel?
    • $\frac{5}{26}$
    • $\frac{7}{52}$
    • $\frac{3}{11}$
    • $\frac{9}{26}$
    Show the answer

    Answer: $\frac{5}{26}$

    Sample space S : Set of English Alphabets
    n(S) = 26

    Let event E represent the set of all vowels.
    n(E) = 5

    Probability(Selecting a vowel from set of English Alphabets) = $\frac{n(E)}{n(S)}$
    = $\frac{5}{26}$

  16. There are 10 boys and 12 girls in a class. If 3 students are selected at random. What is the probability that 1 girl and 2 boys are selected
    • $\frac{3}{11}$
    • $\frac{11}{54}$
    • $\frac{9}{13}$
    • $\frac{27}{77}$
    Show the answer

    Answer: $\frac{27}{77}$

    n(S) : Total number of ways of selecting 3 students from 22 students in the class : $^{22}C_3$

    n(E) : Selecting 2 boys from 10 boys and 1 girl from 12 girls : $^{10}C_2$ x $^{12}C_1$

    P(2 boys and 1 girl) = $\frac{^{10}C_2\text{ x }^{12}C_1}{^{22}C_3}$ = $\frac{540}{1540}$ = $\frac{27}{77}$

  17. What is the probability of getting 2 heads when two coins are tossed together?
    • $\frac{1}{4}$
    • $\frac{3}{4}$
    • $\frac{4}{5}$
    • $\frac{5}{9}$
    Show the answer

    Answer: $\frac{1}{4}$

    Total number of outcomes when a coin is tossed is 2.
    Hence total number of outcomes when two coins are tossed is 2x2 = 4
    Sample Space S - {HH, HT, TH, TT}
    n(S) = 4

    Let E be event of getting heads on both the coins - {HH}
    n(E) = 1

    P(E) = $\frac{n(E)}{n(S)}$
    = $\frac{1}{4}$

  18. What is the probability of getting at most 2 heads when 3 coins are tossed?
    • $\frac{7}{8}$
    • $\frac{3}{7}$
    • $\frac{2}{5}$
    • $\frac{1}{4}$
    Show the answer

    Answer: $\frac{7}{8}$

    Total number of outcomes when a coin is tossed is 2.
    Hence total number of outcomes when 3 coins are tossed is $2^3$ = 8
    Sample Space S - {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
    n(S) = 8

    Let E be event of getting at most 2 heads - {HHT, HTH, HTT, THH, THT, TTH, TTT}
    n(E) = 7

    P(E) = $\frac{n(E)}{n(S)}$ = $\frac{7}{8}$

  19. Four coins are tossed. What is the probability of getting exactly 3 heads?
    • $\frac{1}{2}$
    • $\frac{3}{16}$
    • $\frac{1}{4}$
    • $\frac{2}{9}$
    Show the answer

    Answer: $\frac{1}{4}$

    Total number of outcomes when a coin is tossed is 2.
    Hence total number of outcomes when 4 coins are tossed is $2^4$ = 16
    Sample Space S - {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
    n(S) = 8

    Let E be the event of getting exactly 3 heads when 4 coins are tossed.
    n(E) = Number of ways of getting exactly 3 heads when 4 coins are tossed = $^4C_3$

    P(E) = $\frac{n(E)}{n(S)}$ = $\frac{^4C_3}{16}$ = $\frac{4}{16}$ = $\frac{1}{4}$

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