Probability

www.akankshaclasses.com
आ
CLASS X Mathematics ~4 marks/year Ch 14 of 14
Probability

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • Theoretical (classical) probability measures how likely an event is, assuming every outcome is equally likely: $P(E)=\dfrac{\text{number of outcomes favourable to }E}{\text{number of all possible outcomes}}$.
  • Probability is always a number between $0$ and $1$: $0\le P(E)\le1$. An impossible event has $P=0$; a sure (certain) event has $P=1$.
  • Complement rule: for any event $E$, $P(E)+P(\bar E)=1$, so $P(\bar E)=1-P(E)$ where $\bar E$ means "not $E$".
  • The sum of the probabilities of all elementary events of an experiment is $1$.
  • Board weightage: ~4 marks/year — usually one or two direct problems on coins, dice, cards, balls or number tickets, plus a complement-rule question.
Detailed notes

1. What probability measures

Probability puts a number on how likely something is to happen. The number runs from $0$ (will never happen) to $1$ (sure to happen), with everything uncertain sitting in between. In Class 9 you found probability by experiment — actually tossing a coin many times and counting. That is empirical (experimental) probability:

$$P(E)=\dfrac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$$

This works for coins and dice, but you cannot "repeat" the launch of a satellite or an earthquake thousands of times. So in Class 10 we switch to a method that needs no experiment at all — we just count the possible outcomes and reason. That is theoretical probability.

2. Equally likely outcomes

The whole theory rests on one idea: equally likely outcomes — outcomes that have the same chance of occurring, with no reason to favour one over another.

  • A fair (unbiased) coin tossed at random lands head or tail — the two outcomes are equally likely.
  • A fair die thrown once shows $1,2,3,4,5$ or $6$ — six equally likely outcomes.
  • But not always: a bag with $4$ red balls and $1$ blue ball — drawing "red" and drawing "blue" are not equally likely (red is more likely). However, drawing each individual ball is equally likely.

From here on, this whole chapter assumes all experiments have equally likely outcomes, unless said otherwise.

3. The definition of theoretical probability

The theoretical probability (also called classical probability) of an event $E$, written $P(E)$, is:

$$P(E)=\dfrac{\text{Number of outcomes favourable to }E}{\text{Number of all possible outcomes of the experiment}}$$

where we assume the outcomes are equally likely. "Favourable to $E$" simply means the outcomes that make $E$ happen. (This definition was given by Pierre Simon Laplace in 1795.) We will simply call theoretical probability "probability" from now on.

NCERT Example 1 — a coin tossed once

Tossing a coin once has two possible outcomes: Head (H) and Tail (T). Let $E=$ "getting a head". Favourable outcomes $=1$ (just H). So

$P(E)=P(\text{head})=\dfrac{1}{2}$, and similarly $P(\text{tail})=\dfrac{1}{2}$.

4. Picking one of several equally likely things

NCERT Example 2 — bag with a red, a blue and a yellow ball

Kritika draws one ball without looking, so each of the $3$ balls is equally likely. Number of possible outcomes $=3$.

(i) $P(\text{yellow})=\dfrac{1}{3}$    (ii) $P(\text{red})=\dfrac{1}{3}$    (iii) $P(\text{blue})=\dfrac{1}{3}$.

Elementary event: an event with exactly one outcome (like "yellow", or "head"). A key fact: the probabilities of all the elementary events of an experiment add up to $1$. In Example 1, $P(\text{H})+P(\text{T})=\tfrac12+\tfrac12=1$. In Example 2, $P(Y)+P(R)+P(B)=\tfrac13+\tfrac13+\tfrac13=1$.

5. Events with more than one favourable outcome

NCERT Example 3 — one throw of a die

(i) $E=$ "a number greater than $4$". Possible outcomes $=6$; favourable are $5,6$, so $2$ of them.

$P(E)=\dfrac{2}{6}=\dfrac{1}{3}$.

(ii) $F=$ "a number less than or equal to $4$". Favourable are $1,2,3,4$, so $4$ of them.

$P(F)=\dfrac{4}{6}=\dfrac{2}{3}$.

Events $E$ and $F$ here are not elementary — each has more than one outcome. Notice $P(E)+P(F)=\tfrac13+\tfrac23=1$. That is no accident — it leads straight to the complement rule.

6. Complementary events — the "not E" rule

The event "not $E$" (everything that is not $E$) is called the complement of $E$, written $\bar E$. Together $E$ and $\bar E$ cover all outcomes, so their probabilities add to $1$:

$$P(E)+P(\bar E)=1\qquad\Longrightarrow\qquad P(\bar E)=1-P(E)$$

$E$ and $\bar E$ are called complementary events. This is one of the most useful shortcuts in the chapter: when "the long way" counts a lot of outcomes, count the complement instead and subtract. Example: in Example 3, "greater than $4$" and "less than or equal to $4$" are complementary, which is why their probabilities summed to $1$.

7. Impossible events, sure events, and the range of $P$

Throw a die once and ask two odd questions:

  • Getting an $8$: no face is marked $8$, so favourable outcomes $=0$. Then $P(\text{getting }8)=\dfrac{0}{6}=0$. An event that can never happen is an impossible event, with probability $0$.
  • Getting a number less than $7$: every face ($1$–$6$) qualifies, so all $6$ outcomes are favourable. Then $P=\dfrac{6}{6}=1$. An event sure to happen is a sure (certain) event, with probability $1$.

Since the number of favourable outcomes is always between $0$ and the total, the probability of any event always satisfies:

$$0\le P(E)\le1$$

So a value like $-1.5$ or $\tfrac{7}{5}$ can never be a probability.

8. Playing cards — know the deck

A standard deck has 52 cards in $4$ suits of $13$ each:

  • Black suits: spades ($\spadesuit$) and clubs ($\clubsuit$).
  • Red suits: hearts ($\heartsuit$) and diamonds ($\diamondsuit$).
  • Each suit has: ace, king, queen, jack, $10,9,8,7,6,5,4,3,2$.
  • Face cards: king, queen, jack — so $3$ per suit $=12$ face cards in the deck.
  • $4$ aces, $4$ kings, $4$ queens, $4$ jacks; $26$ red cards and $26$ black cards.
NCERT Example 4 — one card from a well-shuffled deck

Well-shuffling makes all $52$ outcomes equally likely.

(i) $E=$ "the card is an ace". There are $4$ aces, so $P(E)=\dfrac{4}{52}=\dfrac{1}{13}$.

(ii) $F=$ "not an ace". Favourable $=52-4=48$, so $P(F)=\dfrac{48}{52}=\dfrac{12}{13}$.

Shortcut: $F=\bar E$, so $P(F)=1-P(E)=1-\tfrac{1}{13}=\tfrac{12}{13}$ — same answer using the complement rule.

9. Using the complement rule directly

NCERT Example 5 — tennis match

$P(\text{Sangeeta wins})=0.62$. Reshma wins exactly when Sangeeta does not, so the two events are complementary: $P(\text{Reshma wins})=1-0.62=0.38$.

NCERT Example 6 — same / different birthdays (ignoring leap year)

Savita's birthday is fixed on some day. Hamida's birthday can be any of $365$ equally likely days.

(i) Different birthdays: favourable days for Hamida $=365-1=364$, so $P(\text{different})=\dfrac{364}{365}$.

(ii) Same birthday is the complement: $P(\text{same})=1-\dfrac{364}{365}=\dfrac{1}{365}$.

10. More counting problems (students, marbles)

NCERT Example 7 — name card of a girl / boy

$40$ students: $25$ girls, $15$ boys; one name card drawn. Total outcomes $=40$.

(i) $P(\text{girl})=\dfrac{25}{40}=\dfrac{5}{8}$.

(ii) $P(\text{boy})=\dfrac{15}{40}=\dfrac{3}{8}$. Check by complement: $1-\tfrac58=\tfrac38$. $\checkmark$

NCERT Example 8 — box of marbles ($3$ blue, $2$ white, $4$ red)

Total marbles $=3+2+4=9$, all equally likely.

(i) $P(\text{white})=\dfrac{2}{9}$    (ii) $P(\text{blue})=\dfrac{3}{9}=\dfrac{1}{3}$    (iii) $P(\text{red})=\dfrac{4}{9}$.

Note $P(W)+P(B)+P(R)=\tfrac29+\tfrac39+\tfrac49=1$.

11. Two coins together — "at least one head"

NCERT Example 9 — two coins tossed together

List all outcomes as (first coin, second coin): $(H,H),(H,T),(T,H),(T,T)$ — so $4$ equally likely outcomes.

$E=$ "at least one head" means H,H or H,T or T,H — that is $3$ favourable outcomes.

$P(E)=\dfrac{3}{4}$.

By complement: $\bar E=$ "no head" $=(T,T)$, just $1$ outcome, so $P(\bar E)=\tfrac14$ and $P(E)=1-\tfrac14=\tfrac34$.

Tip: the moment you read "at least one", think complement — "at least one head" $=1-P(\text{no head})$. It is almost always faster.

12. Two dice — the $6\times6=36$ table

When two dice (say blue and grey) are thrown together, every outcome is an ordered pair (blue, grey). There are $6\times6=36$ equally likely outcomes. Note $(1,4)$ and $(4,1)$ are different.

NCERT Example 13 — sum of the two dice

(i) Sum $=8$: favourable pairs are $(2,6),(3,5),(4,4),(5,3),(6,2)$ — $5$ of them. So $P=\dfrac{5}{36}$.

(ii) Sum $=13$: impossible (max sum is $12$), so $P=\dfrac{0}{36}=0$.

(iii) Sum $\le 12$: every one of the $36$ outcomes qualifies (sure event), so $P=\dfrac{36}{36}=1$.

13. Geometric probability (length / area) — beyond the exam

Some experiments have infinitely many outcomes (any point on a line, any point in a region). We cannot count them, so we use ratios of lengths or areas. These examples (10, 11) are marked "not from the examination point of view", but the idea is easy and elegant.

NCERT Example 10* — music stops in first half-minute

Music can stop at any point of the line from $0$ to $2$ minutes (total length $2$). Favourable is $0$ to $\tfrac12$ (length $\tfrac12$). So $P(E)=\dfrac{1/2}{2}=\dfrac{1}{4}$.

NCERT Example 11* — helicopter crash inside the lake

Whole region area $=4.5\times9=40.5\ \text{km}^2$; lake area $=2.5\times3=7.5\ \text{km}^2$. So $P=\dfrac{7.5}{40.5}=\dfrac{75}{405}=\dfrac{5}{27}$.

14. Defects — "acceptable to" problems

NCERT Example 12 — carton of 100 shirts

$100$ shirts: $88$ good, $8$ minor defects, $4$ major defects. One drawn at random; $100$ equally likely outcomes.

(i) Jimmy accepts only good shirts: favourable $=88$, so $P=\dfrac{88}{100}=0.88$.

(ii) Sujatha rejects only major-defect shirts (accepts good + minor): favourable $=88+8=96$, so $P=\dfrac{96}{100}=0.96$.

15. NCERT Exercise 14.1 — fully solved (Q1–Q12)

Q1. Complete the statements.

  • (i) $P(E)+P(\text{not }E)=\mathbf{1}$.
  • (ii) The probability of an event that cannot happen is $\mathbf{0}$; such an event is called an impossible event.
  • (iii) The probability of an event certain to happen is $\mathbf{1}$; such an event is called a sure (certain) event.
  • (iv) The sum of the probabilities of all the elementary events of an experiment is $\mathbf{1}$.
  • (v) The probability of an event is greater than or equal to $\mathbf{0}$ and less than or equal to $\mathbf{1}$.

Q2. Which experiments have equally likely outcomes?

  • (i) Car starts / does not start — not equally likely (depends on car condition).
  • (ii) Shot hits / misses — not equally likely (depends on the player's skill).
  • (iii) True-false answer right / wrong — equally likely (a pure guess is 50-50).
  • (iv) Baby is a boy / girl — equally likely.

Q3. Why is tossing a coin fair for a football toss? Because a coin is unbiased: head and tail are equally likely, so neither team is favoured. The result is completely unpredictable and fair.

Q4. Which cannot be a probability? (A) $\tfrac23$ valid, (B) $-1.5$ invalid (negative), (C) $15\%=0.15$ valid, (D) $0.7$ valid. Answer: (B), since probability must lie in $[0,1]$.

Q5. $P(E)=0.05\Rightarrow P(\text{not }E)=1-0.05=\mathbf{0.95}$.

Q6. Bag has lemon candies only.

  • (i) $P(\text{orange})=\mathbf{0}$ — impossible, there are no orange candies.
  • (ii) $P(\text{lemon})=\mathbf{1}$ — sure event, every candy is lemon.

Q7. $P(\text{not same birthday})=0.992$, so $P(\text{same birthday})=1-0.992=\mathbf{0.008}$.

Q8. Bag: $3$ red $+5$ black $=8$ balls. (i) $P(\text{red})=\dfrac{3}{8}$. (ii) $P(\text{not red})=1-\dfrac{3}{8}=\dfrac{5}{8}$.

Q9. Box: $5$ red, $8$ white, $4$ green; total $=17$. (i) $P(\text{red})=\dfrac{5}{17}$. (ii) $P(\text{white})=\dfrac{8}{17}$. (iii) $P(\text{not green})=1-\dfrac{4}{17}=\dfrac{13}{17}$.

Q10. Coins: $100$ of 50p, $50$ of $\text{Rs }1$, $20$ of $\text{Rs }2$, $10$ of $\text{Rs }5$; total $=180$. (i) $P(\text{50p})=\dfrac{100}{180}=\dfrac{5}{9}$. (ii) $P(\text{not Rs }5)=1-\dfrac{10}{180}=\dfrac{170}{180}=\dfrac{17}{18}$.

Q11. Tank: $5$ male $+8$ female $=13$ fish. $P(\text{male})=\dfrac{5}{13}$.

Q12. Spinner with $1$–$8$, all equally likely (total $=8$).

  • (i) $P(8)=\dfrac{1}{8}$.
  • (ii) Odd numbers $1,3,5,7$: $P=\dfrac{4}{8}=\dfrac{1}{2}$.
  • (iii) Greater than $2$: $3,4,5,6,7,8$ — $6$ values, $P=\dfrac{6}{8}=\dfrac{3}{4}$.
  • (iv) Less than $9$: all $8$ qualify, $P=\dfrac{8}{8}=1$ (sure event).

16. NCERT Exercise 14.1 — fully solved (Q13–Q25)

Q13. A die thrown once.

  • (i) Prime numbers $2,3,5$: $P=\dfrac{3}{6}=\dfrac{1}{2}$.
  • (ii) Number between $2$ and $6$ ($3,4,5$): $P=\dfrac{3}{6}=\dfrac{1}{2}$.
  • (iii) Odd numbers $1,3,5$: $P=\dfrac{3}{6}=\dfrac{1}{2}$.

Q14. One card from $52$.

  • (i) King of red colour: $2$ (king of hearts, king of diamonds), $P=\dfrac{2}{52}=\dfrac{1}{26}$.
  • (ii) A face card: $12$ in all, $P=\dfrac{12}{52}=\dfrac{3}{13}$.
  • (iii) A red face card: $6$ (hearts + diamonds: K,Q,J each), $P=\dfrac{6}{52}=\dfrac{3}{26}$.
  • (iv) The jack of hearts: just $1$, $P=\dfrac{1}{52}$.
  • (v) A spade: $13$, $P=\dfrac{13}{52}=\dfrac{1}{4}$.
  • (vi) The queen of diamonds: just $1$, $P=\dfrac{1}{52}$.

Q15. Five diamond cards: 10, J, Q, K, A.

  • (i) $P(\text{queen})=\dfrac{1}{5}$.
  • (ii) Queen removed, $4$ cards left. (a) $P(\text{ace})=\dfrac{1}{4}$. (b) $P(\text{queen})=\dfrac{0}{4}=0$ (no queen left — impossible).

Q16. $12$ defective $+132$ good $=144$ pens. $P(\text{good})=\dfrac{132}{144}=\dfrac{11}{12}$.

Q17. Lot of $20$ bulbs, $4$ defective.

  • (i) $P(\text{defective})=\dfrac{4}{20}=\dfrac{1}{5}$.
  • (ii) One non-defective bulb removed and not replaced, so $19$ left with $4$ still defective ($15$ good). $P(\text{not defective})=\dfrac{15}{19}$.

Q18. $90$ discs numbered $1$–$90$.

  • (i) Two-digit numbers are $10$ to $90$, that is $81$ numbers, $P=\dfrac{81}{90}=\dfrac{9}{10}$.
  • (ii) Perfect squares up to $90$: $1,4,9,16,25,36,49,64,81$ — $9$ numbers, $P=\dfrac{9}{90}=\dfrac{1}{10}$.
  • (iii) Divisible by $5$: $5,10,\dots,90$ — $18$ numbers, $P=\dfrac{18}{90}=\dfrac{1}{5}$.

Q19. Die with faces A, B, C, D, E, A. (i) $P(\text{A})=\dfrac{2}{6}=\dfrac{1}{3}$ (A appears twice). (ii) $P(\text{D})=\dfrac{1}{6}$.

Q20*. Die dropped on a $3\text{ m}\times2\text{ m}$ rectangle, circle of diameter $1$ m (radius $\tfrac12$ m). Rectangle area $=6\ \text{m}^2$; circle area $=\pi r^2=\pi\left(\tfrac12\right)^2=\dfrac{\pi}{4}\ \text{m}^2$. So $P=\dfrac{\pi/4}{6}=\dfrac{\pi}{24}$.

Q21. $144$ pens, $20$ defective, $124$ good. (i) $P(\text{buys})=P(\text{good})=\dfrac{124}{144}=\dfrac{31}{36}$. (ii) $P(\text{will not buy})=\dfrac{20}{144}=\dfrac{5}{36}$ (or $1-\tfrac{31}{36}$).

Q22. Sum on two dice ($36$ outcomes).

  • (i) Counts of favourable pairs: sum $2{:}1,\ 3{:}2,\ 4{:}3,\ 5{:}4,\ 6{:}5,\ 7{:}6,\ 8{:}5,\ 9{:}4,\ 10{:}3,\ 11{:}2,\ 12{:}1$. So the probabilities are $\dfrac{1}{36},\dfrac{2}{36},\dfrac{3}{36},\dfrac{4}{36},\dfrac{5}{36},\dfrac{6}{36},\dfrac{5}{36},\dfrac{4}{36},\dfrac{3}{36},\dfrac{2}{36},\dfrac{1}{36}$.
  • (ii) The argument is wrong. The $11$ sums are not equally likely (e.g. sum $7$ has $6$ outcomes but sum $2$ has only $1$), so each cannot have probability $\tfrac{1}{11}$.

Q23. Toss a coin $3$ times: $8$ equally likely outcomes (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT). Hanif wins only with HHH or TTT — $2$ outcomes. $P(\text{wins})=\dfrac{2}{8}=\dfrac{1}{4}$, so $P(\text{loses})=1-\dfrac{1}{4}=\dfrac{3}{4}$.

Q24. A die thrown twice ($36$ outcomes).

  • (i) $5$ comes up neither time: outcomes with no $5$ on either die $=5\times5=25$, so $P=\dfrac{25}{36}$.
  • (ii) $5$ comes up at least once: complement of (i), $P=1-\dfrac{25}{36}=\dfrac{11}{36}$.

Q25. Correct or not?

  • (i) Incorrect. Two coins give outcomes HH, HT, TH, TT — "one of each" covers HT and TH ($2$ outcomes), so the three cases are not equally likely; each is not $\tfrac13$. ($P(2\text{ heads})=\tfrac14$, $P(2\text{ tails})=\tfrac14$, $P(\text{one of each})=\tfrac24$.)
  • (ii) Correct. A die has $3$ odd and $3$ even numbers — equally likely — so $P(\text{odd})=\dfrac{3}{6}=\dfrac{1}{2}$.

17. Common mistakes to avoid

  • Assuming outcomes are equally likely when they are not (e.g. "car starts / does not start").
  • Giving a probability outside $[0,1]$ — like $-1.5$ or $\tfrac{7}{5}$. Always check $0\le P\le1$.
  • Forgetting that two coins give four outcomes (HH, HT, TH, TT), not three — and two dice give $36$, not $12$.
  • Treating $(1,4)$ and $(4,1)$ as the same outcome on two dice — they are different.
  • Forgetting "not replaced": the total drops by $1$ for the second draw (see Q17).
  • Miscounting face cards ($12$ total, $6$ red) or red cards ($26$).
  • Not simplifying the final fraction.

18. Quick revision checklist

  • $P(E)=\dfrac{\text{favourable outcomes}}{\text{all outcomes}}$, assuming equally likely outcomes.
  • $0\le P(E)\le1$; impossible $\Rightarrow0$, sure $\Rightarrow1$.
  • $P(E)+P(\bar E)=1$; use the complement for "at least one" and "not" questions.
  • Sum of probabilities of all elementary events $=1$.
  • One die: $6$ outcomes. Two coins: $4$. Two dice: $36$. Deck: $52$ ($4$ aces, $12$ face, $26$ red).
Practice MCQs
1. The probability of a sure event is:
  1. $0$
  2. $\dfrac{1}{2}$
  3. $1$
  4. $-1$
Answer: (C) A certain event always happens, so $P=1$.
2. Which of the following cannot be the probability of an event?
  1. $0.7$
  2. $\dfrac{2}{3}$
  3. $-0.4$
  4. $0$
Answer: (C) Probability cannot be negative; it must lie in $[0,1]$.
3. A die is thrown once. The probability of getting an even number is:
  1. $\dfrac{1}{6}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{3}$
  4. $\dfrac{2}{3}$
Answer: (B) Even numbers $2,4,6$ give $\dfrac{3}{6}=\dfrac{1}{2}$.
4. One card is drawn from a deck of $52$. The probability that it is a king is:
  1. $\dfrac{1}{13}$
  2. $\dfrac{1}{26}$
  3. $\dfrac{1}{4}$
  4. $\dfrac{4}{13}$
Answer: (A) $4$ kings, so $\dfrac{4}{52}=\dfrac{1}{13}$.
5. If $P(E)=0.35$, then $P(\bar E)$ equals:
  1. $0.35$
  2. $0.65$
  3. $0.55$
  4. $1.35$
Answer: (B) $P(\bar E)=1-0.35=0.65$.
6. Two coins are tossed together. The probability of getting at least one head is:
  1. $\dfrac{1}{4}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{3}{4}$
  4. $1$
Answer: (C) Outcomes HH, HT, TH favour it ($3$ of $4$), so $\dfrac{3}{4}$.
7. A bag has $3$ red and $2$ blue balls. The probability of drawing a blue ball is:
  1. $\dfrac{2}{5}$
  2. $\dfrac{3}{5}$
  3. $\dfrac{2}{3}$
  4. $\dfrac{1}{5}$
Answer: (A) $\dfrac{2}{2+3}=\dfrac{2}{5}$.
8. The probability of getting a number greater than $6$ on a single throw of a die is:
  1. $1$
  2. $\dfrac{1}{6}$
  3. $0$
  4. $\dfrac{1}{2}$
Answer: (C) No face exceeds $6$ — impossible event, $P=0$.
9. A card is drawn from $52$ cards. The probability it is a face card is:
  1. $\dfrac{3}{13}$
  2. $\dfrac{1}{13}$
  3. $\dfrac{1}{4}$
  4. $\dfrac{4}{13}$
Answer: (A) $12$ face cards, so $\dfrac{12}{52}=\dfrac{3}{13}$.
10. Two dice are thrown. The probability that the sum is $7$ is:
  1. $\dfrac{1}{6}$
  2. $\dfrac{5}{36}$
  3. $\dfrac{1}{12}$
  4. $\dfrac{7}{36}$
Answer: (A) Pairs $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — $6$ of $36=\dfrac{1}{6}$.
Assertion–Reason
A: The probability of an impossible event is $0$.   R: For an impossible event, the number of favourable outcomes is $0$.
Answer: Both A and R are true, and R is the correct explanation of A — with $0$ favourable outcomes, $P=\dfrac{0}{\text{total}}=0$.
A: $P(E)+P(\bar E)=1$ for every event $E$.   R: Probability of an event can be greater than $1$.
Answer: A is true, R is false — probability always satisfies $0\le P(E)\le1$, so it can never exceed $1$.
Previous-year questions
Q1. A die is thrown once. Find the probability of getting a prime number. (CBSE, 1 mark)
Answer: Primes are $2,3,5$, so $P=\dfrac{3}{6}=\dfrac{1}{2}$.
Q2. A card is drawn at random from a well-shuffled deck of $52$ cards. Find the probability of getting (i) a red face card, (ii) a card that is neither a king nor a queen. (CBSE, 2 marks)
Answer: (i) Red face cards $=6$, so $P=\dfrac{6}{52}=\dfrac{3}{26}$. (ii) Kings + queens $=8$, so neither $=52-8=44$, $P=\dfrac{44}{52}=\dfrac{11}{13}$.
Q3. A bag contains $5$ red, $4$ green and $3$ black balls. One ball is drawn at random. Find the probability that it is (i) not black, (ii) red or green. (CBSE, 3 marks)
Answer: Total $=12$. (i) $P(\text{not black})=1-\dfrac{3}{12}=\dfrac{9}{12}=\dfrac{3}{4}$. (ii) Red or green $=5+4=9$, so $P=\dfrac{9}{12}=\dfrac{3}{4}$.
Q4. Two dice are thrown simultaneously. Find the probability of getting (i) a doublet, (ii) a sum of $9$. (CBSE, 3 marks)
Answer: Total $=36$. (i) Doublets $(1,1),(2,2),\dots,(6,6)$ — $6$ of them, $P=\dfrac{6}{36}=\dfrac{1}{6}$. (ii) Sum $9$: $(3,6),(4,5),(5,4),(6,3)$ — $4$ of them, $P=\dfrac{4}{36}=\dfrac{1}{9}$.
Want personal coaching in Dwarka?
Book a free demo class
More Class 10 Mathematics chapters
Real Numbers · Polynomials · Pair of Linear Equations in Two Variables · Quadratic Equations · Arithmetic Progressions · Triangles · Coordinate Geometry · Introduction to Trigonometry