Unit 13: Probability

Review Exercise 13 Solutions

Step-by-step solved textbook exercises for MCQ review questions, definitions, coins, cards, colored balls, and lots of defective items.

Question 1 Multiple Choice
Choose the correct option.

(All 10 textbook MCQs are fully answered in the table below. Play the interactive quiz in the sandbox at the bottom to test your knowledge!)

Solution
Sub-Q Question Statement Correct Option Mathematical Explanation
i Each element of the sample space is called: (c) Sample point Every individual element inside a sample space set represents a single outcome, mathematically known as a sample point.
ii An outcome which represents how many times we expect the things to happen is called: (a) Outcomes Textbook Key matches (a). Note that in mathematics, the term representing expected occurrences is actually expected frequency, but the textbook key lists Outcomes.
iii Which one tells us how often a specific event occurs relative to the total number of trials? (c) Relative frequency Relative frequency is the ratio of the frequency of an event to the total number of repetitions/trials.
iv Estimated probability of an event occurring is also known as: (a) Relative frequency The empirical probability is defined as the relative frequency of the event in the long run.
v The sum of all expected frequencies is equal to the fixed number of: (a) Trials The sum of expected frequencies is \(\sum E_i = \sum (N \times P_i) = N \sum P_i = N \times 1 = N\) (total trials).
vi The chance of occurrence of a particular event is called: (c) Probability Probability is the numerical value that measures the likelihood of an event occurring.
vii An event which will probably occur. It has greater chance to occur is called: (b) Likely event An event with a probability \(P(A) > 0.5\) is considered physically likely to occur.
viii Find the total number of possible sample space when 4 dice are rolled: (d) 64 For k dice, each having 6 faces, the sample space size is 6k. For 4 dice, 64 = 1296.
ix While rolling a pair of dice, what will be the probability of double 2? (c) 136 Double 2 is a single outcome {(2, 2)} out of 36 total outcomes, so P = 136.
x A card is chosen from a pack of 52 playing cards, find the probability of getting no jack and king: (c) 1113 Jacks = 4, Kings = 4 (8 cards total). Favorable cards = 52 − 8 = 44. P = 4452 = 1113.
Question 2 Definitions
Define the following:
  • (i) Relative frequency
  • (ii) Expected frequency
Solution
  • (i) Relative Frequency:
    Relative frequency tells us how often a specific event occurs relative to the total number of trials or events. It is computed as:

    Relative Frequency = Frequency of specific event (x)Total frequency (N) = xΣf

  • (ii) Expected Frequency:
    Expected frequency is an estimated measure that calculates how often an event is theoretically expected to occur based on its probability distribution. It is computed as:

    Expected Frequency = Total number of trials (N) × Probability of the event (P(A)) = N × P(A)

Question 3 Urn Drawing
An urn contains 10 red balls, 5 green balls and 8 blue balls. Find the probability of selecting at random:
  • (i) a green ball
  • (ii) a red ball
  • (iii) a blue ball
  • (iv) not a red ball
  • (v) not a green ball
Solution

First, list the total count of each color:

  • Red balls: n(R) = 10
  • Green balls: n(G) = 5
  • Blue balls: n(B) = 8

Total outcomes in the sample space:

n(S) = 10 + 5 + 8 = 23

  • (i) Probability of selecting a green ball:
    P(G) = n(G)n(S) = 523.
  • (ii) Probability of selecting a red ball:
    P(R) = n(R)n(S) = 1023.
  • (iii) Probability of selecting a blue ball:
    P(B_ball) = n(B)n(S) = 823.
  • (iv) Probability of selecting not a red ball:
    This is the complement of event R:
    P(R') = 1 − P(R) = 1 − 1023 = 1323.
  • (v) Probability of selecting not a green ball:
    This is the complement of event G:
    P(G') = 1 − P(G) = 1 − 523 = 1823.
Question 4 Correction Applied
Three coins are tossed together. What is the probability of getting:
  • (i) exactly three heads
  • (ii) at least two tails
  • (iii) not at least two heads
  • (iv) not exactly two heads
Solution

When three coins are tossed together, the sample space S contains \(2^3 = 8\) outcomes:

S = {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT},   n(S) = 8

  • (i) Exactly three heads:
    Let A be this event: A = {HHH}, n(A) = 1.
    P(A) = 18.
  • (ii) At least two tails (meaning 2 or 3 tails):
    Let B be this event: B = {TTT, TTH, THT, HTT}, n(B) = 4.
    P(B) = 48 = 12.
  • (iii) Not at least two heads:
    Let C be the event of getting at least two heads: C = {HHH, HHT, HTH, THH}, n(C) = 4.
    The probability of at least two heads is: P(C) = 48 = 12.
    The probability of NOT getting at least two heads is:
    P(C') = 1 − P(C) = 1 − 12 = 12.
  • (iv) Not exactly two heads:
    Let D be the event of getting exactly two heads: D = {HHT, HTH, THH}, n(D) = 3.
    The probability of exactly two heads is: P(D) = 38.
    The probability of NOT getting exactly two heads is the complement:
    P(D') = 1 − P(D) = 1 − 38 = 58.
Mathematical Note on Textbook Typo In part (iv) of Q.4, the textbook intermediate step writes the subtraction line incorrectly as 1 − 18 = 58 (writing 18 instead of 38). The final answer in the textbook (58) is correct because 1 − 38 = 58. We corrected this typo to show the correct intermediate math.
Question 5 Playing Cards
A card is drawn from a well shuffled pack of 52 playing cards. What will be the probability of getting:
  • (i) King or Jack of red color
  • (ii) Not "2" of club and spade
Solution

Total cards in the pack: n(S) = 52

  • (i) King or Jack of red color:
    Red kings = 2 (Hearts, Diamonds). Red jacks = 2 (Hearts, Diamonds).
    Let A be this event: n(A) = 2 + 2 = 4 cards.
    P(A) = 452 = 113.
  • (ii) Not "2" of club and spade (meaning not 2 of club and not 2 of spade):
    Let B be the event of getting the "2" of club or spade. There are 2 such cards in a deck:
    n(B) = 2 cards.
    The probability of selecting one of these cards is:
    P(B) = 252 = 126.
    The probability of NOT selecting these cards is:
    P(B') = 1 − P(B) = 1 − 126 = 2526.
Question 6 Relative Frequency
Six coins are tossed 600 times. The number of occurrence of tails are recorded and shown in the table given below.
No. of tails 0 1 2 3 4 5 6
Frequency 110 90 105 80 76 123 16
Find the relative frequency of the given table.
Solution

The total trials frequency (Σf) = 600.

We divide each frequency by 600 to find the relative frequencies:

No. of Tails Frequency (f) Relative Frequency calculation Relative Frequency (r.f.)
0 110 110600 1160 ≈ 0.183
1 90 90600 320 = 0.150
2 105 105600 740 = 0.175
3 80 80600 215 ≈ 0.133
4 76 76600 19150 ≈ 0.127
5 123 123600 41200 = 0.205
6 16 16600 275 ≈ 0.027
Total Σf = 600 — 1.000
Question 7 Defective Lot
From a lot containing 25 items, 8 items are defective. Find the relative frequency of non-defective items, also find the expected frequency of non-defective items.
Solution

We are given:

  • Total items in the lot (N) = 25
  • Defective items = 8

Step 1: Find the relative frequency of non-defective items:

Number of non-defective items = Total items − Defective items = 25 − 8 = 17 items.

Relative Frequency of non-defectives = 1725 = 0.68 (or 68%)

Step 2: Find the expected frequency of non-defective items:

Expected frequency is computed as:

Expected Frequency = N × Relative Frequency = 25 × 1725 = 17

Thus, the relative frequency of non-defective items is 0.68 and the expected frequency is 17.

🎮 Unit 13 Interactive Learning Sandbox

Test your knowledge with an interactive self-grading quiz, or run a physics Galton Board simulation to explore probability distributions!

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