This MCQ module is based on: Problems on Single Events in Probability
TOPIC 35 OF 35
Problems on Single Events in Probability
🎓 Class 10
Mathematics
CBSE
Theory
Ch 14 — Probability
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Problems on Single Events in Probability
Targeting Class 10 level in Probability, with Intermediate difficulty.
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14.1 (continued) — Worked Problems on Single Events
We now build problem-solving fluency with the classical definition \(\;P(E)=\dfrac{\text{favourable outcomes}}{\text{total outcomes}}\) by applying it to a variety of standard situations: coins, dice, cards, marbles in a bag, spinners, and lotteries. For each problem we use the three-step method: identify the sample space, count favourable outcomes, divide.
Example 1 — Bag of Balls
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
Solution. Total balls = 8, so \(n=8\).
(i) Red balls = 3. \(P(\text{red})=\dfrac{3}{8}\).
(ii) \(P(\text{not red})=1-P(\text{red})=1-\dfrac{3}{8}=\dfrac{5}{8}\).
(ii) \(P(\text{not red})=1-P(\text{red})=1-\dfrac{3}{8}=\dfrac{5}{8}\).
Example 2 — Marbles in a Box
There are 5 red marbles, 8 white marbles and 4 green marbles in a box. One marble is taken out of the box at random. What is the probability that the marble is (i) red? (ii) white? (iii) not green?
Solution. Total = 5 + 8 + 4 = 17.
(i) \(P(\text{red})=5/17\) (ii) \(P(\text{white})=8/17\)
(iii) \(P(\text{not green})=1-P(\text{green})=1-4/17=13/17\)
(iii) \(P(\text{not green})=1-P(\text{green})=1-4/17=13/17\)
Example 3 — Defective Pens
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random. What is the probability that Nuri will buy the pen?
Solution. Total pens = 144; good = 124. So \(P(\text{Nuri buys})=P(\text{good})=\dfrac{124}{144}=\dfrac{31}{36}\). And \(P(\text{does not buy})=\dfrac{20}{144}=\dfrac{5}{36}\).
Example 4 — Random Selection from a Class
There are 40 students in Class X of a school of which 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical, then shuffles them and picks one. What is the probability the name written on the card is the name of (i) a girl? (ii) a boy?
(i) \(P(\text{girl})=25/40=5/8\) (ii) \(P(\text{boy})=15/40=3/8\)
Example 5 — Discs Numbered 1–90
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number, (ii) a perfect square, (iii) a number divisible by 5.
(i) Two-digit numbers in 1–90: from 10 to 90 → 81 numbers. \(P=81/90=9/10\).
(ii) Perfect squares ≤ 90: 1, 4, 9, 16, 25, 36, 49, 64, 81 → 9 numbers. \(P=9/90=1/10\).
(iii) Multiples of 5 from 1 to 90: 5, 10, …, 90 → 18 numbers. \(P=18/90=1/5\).
(ii) Perfect squares ≤ 90: 1, 4, 9, 16, 25, 36, 49, 64, 81 → 9 numbers. \(P=9/90=1/10\).
(iii) Multiples of 5 from 1 to 90: 5, 10, …, 90 → 18 numbers. \(P=18/90=1/5\).
Example 6 — Spinner Game
A game consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (Fig. below) and these are equally likely outcomes. What is the probability that it will point at (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
Each sector has equal area → each number is equally likely: probability 1/8.
Total outcomes = 8.
(i) \(P(8) = 1/8\)
(ii) Odd numbers: 1, 3, 5, 7 → \(P=4/8=1/2\)
(iii) Numbers > 2: 3, 4, 5, 6, 7, 8 → \(P=6/8=3/4\)
(iv) Numbers < 9: all 8 outcomes → \(P=8/8=1\) (sure event)
(i) \(P(8) = 1/8\)
(ii) Odd numbers: 1, 3, 5, 7 → \(P=4/8=1/2\)
(iii) Numbers > 2: 3, 4, 5, 6, 7, 8 → \(P=6/8=3/4\)
(iv) Numbers < 9: all 8 outcomes → \(P=8/8=1\) (sure event)
Example 7 — Savita and Hamida Kings
Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (Ignoring a leap year.)
Solution. There are 365 possible days. Fix Savita's birthday. Total possible days for Hamida = 365. For different birthdays, Hamida can be born on any of the remaining 364 days.
(i) \(P(\text{different})=\dfrac{364}{365}\) (ii) \(P(\text{same})=1-\dfrac{364}{365}=\dfrac{1}{365}\)
Example 8 — Rolling Two Dice
Two dice, one blue and one grey, are thrown at the same time. The 36 equally-likely outcomes are listed in the grid below. Find the probability that the sum of the two numbers is (i) 8, (ii) 13, (iii) ≤ 12.
36 equally-likely outcomes. Cells in orange are the 5 outcomes where the two numbers add to 8.
(i) Sum 8 → (2,6),(3,5),(4,4),(5,3),(6,2) = 5 outcomes. \(P(\text{sum}=8)=5/36\).
(ii) Sum 13: impossible (max sum = 12). \(P=0/36=0\).
(iii) Sum ≤ 12: always true. \(P=36/36=1\) (sure event).
(ii) Sum 13: impossible (max sum = 12). \(P=0/36=0\).
(iii) Sum ≤ 12: always true. \(P=36/36=1\) (sure event).
Example 9 — Defective Bulbs
A carton of 24 bulbs contain 6 defective bulbs. One bulb is drawn at random. What is the probability that the bulb is not defective? If the drawn bulb is not defective and is not replaced, and a second bulb is drawn at random, what is the probability the second bulb is not defective?
Initial good bulbs = 18, total = 24. \(P(\text{good on 1st})=18/24=3/4\).
After one good bulb is removed: 17 good, 23 total. \(P(\text{good on 2nd})=17/23\).
After one good bulb is removed: 17 good, 23 total. \(P(\text{good on 2nd})=17/23\).
Example 10 — Alphabet Letters
The letters of the English alphabet are written on 26 identical cards and placed in a box. A card is drawn at random. What is the probability that the card drawn shows (i) a vowel (ii) a consonant (iii) a letter from the word "INDIA" (repeated letters counted once)?
(i) Vowels A, E, I, O, U → 5. \(P=5/26\).
(ii) Consonants = 21. \(P=21/26\).
(iii) Distinct letters of INDIA: I, N, D, A → 4. \(P=4/26=2/13\).
(ii) Consonants = 21. \(P=21/26\).
(iii) Distinct letters of INDIA: I, N, D, A → 4. \(P=4/26=2/13\).
In-text: When an event E has \(P(E)=1\), the event is called a sure event; when \(P(E)=0\), it is called an impossible event. All other events are "uncertain" and lie strictly between 0 and 1.
Example 11 — Lost a Date
Savita asks her friend Hamida about the birthday of Harpreet Singh. Hamida is not sure; she says Harpreet's birth month is April. What is the probability that Harpreet was born on the 29th of April?
Solution. April has 30 days. Each day is equally likely. \(P=\dfrac{1}{30}\).
Activity — Die vs Empirical Probability
L3 Apply
Materials: 1 fair die, tally sheet.
Predict: If a die is rolled 60 times, roughly how many times should each face appear?
- Roll the die 60 times. Tally the outcomes in a table.
- Compute the empirical probability for each face (tally / 60).
- Compare with the theoretical probability 1/6 ≈ 0.167.
- Now roll another 60 times, for 120 total. Recompute. Which gets closer to 1/6?
The empirical ratio for each face stabilises near 1/6 as the number of rolls grows. Any systematic deviation (e.g., 6 appearing almost twice as often as 1) would suggest the die is not fair.
Competency-Based Questions
Scenario: A box contains 12 balls of which some are red. If 6 more red balls are added to the box, the probability of drawing a red ball is doubled. Let the original number of red balls be \(x\).
Q1. Set up and solve an equation for \(x\).
L3 ApplyOriginal P(red) = \(x/12\). New box has \(x+6\) red balls out of 18 total, so new P = \((x+6)/18\). "Doubled" gives \((x+6)/18 = 2\cdot(x/12) = x/6\). Cross-multiply: \(6(x+6)=18x\Rightarrow 6x+36=18x\Rightarrow 12x=36\Rightarrow x=3\). So 3 red balls originally.
Q2. Analyse: In a lottery, 1000 tickets are sold and 1 winner is drawn. Which is more probable — buying 1 ticket out of 1000 or 1 ticket out of 500 when 1 prize is awarded in each lottery? Justify.
L4 AnalyseP(win) in first = 1/1000 = 0.001; in second = 1/500 = 0.002. Buying the same single ticket in the smaller (500-ticket) lottery is twice as likely to win. Probability depends on the number of equally-likely outcomes, not on the size of the prize.
Q3. Evaluate: Kritika draws one card from a deck of 52 and says, "Since there are four 4 suits, the probability of a heart is 1/4". Is her reasoning correct? Would the conclusion still be valid if 3 hearts were missing from the deck?
L5 EvaluateHer reasoning is correct only for a complete deck: 13 hearts out of 52 = 1/4. If 3 hearts are removed, hearts = 10 out of 49 total: \(P=10/49\approx 0.204 \ne 0.25\). The "four suits → 1/4" short-cut silently assumes every suit has equal count, which fails when cards are missing.
Q4. Create: Design a spinner or a bag-of-balls experiment with exactly 12 equally likely outcomes in which one event has probability 1/4, another has probability 1/6, and the two events share no favourable outcomes. Describe the sample space and the two events.
L6 CreateBag with 12 balls labelled 1–12. Event A = {1, 2, 3} → P(A) = 3/12 = 1/4. Event B = {4, 5} → P(B) = 2/12 = 1/6. A ∩ B = ∅ (no shared outcomes). Many valid constructions: any disjoint subsets of sizes 3 and 2 from a 12-outcome sample space work.
Assertion–Reason Questions
Assertion (A): When two dice are rolled, P(sum = 7) > P(sum = 2).
Reason (R): Sum = 7 can occur in 6 different ways, whereas sum = 2 can occur in only 1 way.
Reason (R): Sum = 7 can occur in 6 different ways, whereas sum = 2 can occur in only 1 way.
(a) — P(7) = 6/36 = 1/6; P(2) = 1/36. The counts in R directly justify the inequality in A.
Assertion (A): P(drawing a king from a well-shuffled deck) = 4/52.
Reason (R): The favourable outcomes are the 4 kings and the total outcomes are the 52 cards in the deck.
Reason (R): The favourable outcomes are the 4 kings and the total outcomes are the 52 cards in the deck.
(a) — direct application of the classical definition.
Assertion (A): A bag has 3 red and 5 white balls. The probability of NOT drawing a red is 5/8.
Reason (R): P(not red) = P(white) because red and white are the only colours and any ball is either red or white but not both.
Reason (R): P(not red) = P(white) because red and white are the only colours and any ball is either red or white but not both.
(a) — P(not red) = 5/8 = P(white); R correctly explains the complement reasoning.
Summary of Chapter 14
Key Take-aways
- A random experiment has outcomes that cannot be predicted; the set of all outcomes is the sample space.
- If outcomes are equally likely, \(P(E)=\dfrac{\text{favourable outcomes}}{\text{total outcomes}}\).
- \(0\le P(E)\le 1\); the impossible event has probability 0, the sure event has probability 1.
- Complementary events satisfy \(P(\bar{E})=1-P(E)\).
- The experimental (empirical) probability approaches the theoretical probability as the number of trials grows large — the Law of Large Numbers.
Frequently Asked Questions
How do you find the probability of getting a head in a coin toss?
A fair coin has sample space {H, T} with 2 equally likely outcomes. P(Head) = 1/2 and P(Tail) = 1/2.
What is the probability of getting a prime number on a die?
Sample space = {1,2,3,4,5,6}. Primes are {2,3,5}, i.e. 3 favourable outcomes. P(prime) = 3/6 = 1/2.
How many cards are there in a standard deck and how is it arranged?
A standard deck has 52 cards: 4 suits (spades, hearts, diamonds, clubs) of 13 cards each (Ace, 2-10, Jack, Queen, King). Jacks, Queens and Kings are face cards (12 total); red cards (hearts + diamonds) total 26.
What is the probability of drawing a king from a well-shuffled deck?
There are 4 kings in 52 cards, so P(King) = 4/52 = 1/13.
If a bag has 3 red and 5 black balls, what is P(red)?
Total balls = 8, favourable (red) = 3, so P(red) = 3/8 and P(black) = 5/8.
How do 'not' and 'at least' questions use complementary probability?
Use P(not E) = 1 - P(E). For example, if P(defective) = 1/5, then P(not defective) = 4/5. This trick avoids long case counting.
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