Probability | Mathematics Class 10

Q1

NUMERIC3 marks

Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = ______________.
(ii) The probability of an event that cannot happen is called __________________. Such an event is called ___________.
(iii) The probability of an event that is certain to happen is _______________. Such an event is called ____________.
(iv) The sum of the probabilities of all the elementary events of an experiment is ___________________.
(v) The probability of an event is greater than or equal to ________________ and less than or equal to _______________ .

Basic Theory of Probability

In probability, we study the likelihood of occurrence of events.

An event is a possible outcome or a set of outcomes of an experiment.
The probability of an event E is denoted as \( P(E) \).
For any event: \[ 0 \leq P(E) \leq 1 \]
The complement of an event E is ‘not E’, denoted by \( E' \).
\[ P(E) + P(E') = 1 \]

Solution Roadmap

Recall basic definitions of probability.
Use standard results like complement rule.
Identify types of events: impossible, sure, elementary.
Apply known probability limits.

Solution

(i) Using complement rule:

\[ P(E) + P(E') = 1 \]

Therefore, Probability of an event E + Probability of ‘not E’ = 1.

(ii) An event that cannot occur has no favorable outcomes.

\[ P(\text{impossible event}) = 0 \]

Hence, it is called 0 and such an event is called an impossible event.

(iii) An event that always occurs has probability:

\[ P(\text{sure event}) = 1 \]

Hence, it is called 1 and such an event is called a sure (certain) event.

(iv) Let total sample space be S.

Sum of probabilities of all elementary events in S:

\[ \sum P(\text{all elementary events}) = 1 \]

Therefore, the answer is 1.

(v) For any event E:

\[ 0 \leq P(E) \leq 1 \]

Hence, probability lies between 0 and 1.

Exam Significance

These are fundamental identities of probability frequently asked in CBSE board exams.
Direct 1-mark or fill-in-the-blank questions are common.
Concepts like complement rule are heavily used in higher problems.
Important for competitive exams like NTSE, SSC, Banking, CUET.

Theory: Equally Likely Outcomes

Two or more outcomes are said to be equally likely if each outcome has the same chance of occurring.

For example, when a fair coin is tossed: \[ P(\text{Head}) = P(\text{Tail}) = \frac{1}{2} \]

Solution Roadmap

Check whether outcomes depend on external factors.
If probability varies → not equally likely.
If no bias or influencing factor → equally likely.
Apply standard school-level assumptions where required.

Solution

(i) A driver attempts to start a car

The result depends on multiple real-life factors such as engine condition, fuel, battery, and maintenance.

Therefore, the outcomes "car starts" and "car does not start" do not have equal chances.

Conclusion: Not equally likely.

(ii) A player attempts to shoot a basketball

The result depends on skill level, distance, practice, and external conditions.

Hence, "shoot" and "miss" do not have equal probability.

Conclusion: Not equally likely.

(iii) A trial is made to answer a true–false question

If a student guesses without any knowledge, there are only two possible outcomes:

\[ \text{Right or Wrong} \]

Both outcomes are equally possible under random guessing:

\[ P(\text{Right}) = P(\text{Wrong}) = \frac{1}{2} \]

Conclusion: Equally likely.

(iv) A baby is born

In elementary probability, we assume:

\[ P(\text{Boy}) = P(\text{Girl}) = \frac{1}{2} \]

Although real-world data may slightly vary, this assumption is valid at the school level.

Conclusion: Equally likely.

Final Answer: Only (iii) and (iv) have equally likely outcomes.

Exam Significance

Tests conceptual clarity rather than calculation.
Frequently asked as reason-based questions in CBSE exams.
Helps in understanding assumptions used in probability problems.
Important for competitive exams like NTSE, CUET, SSC.

Theory: Fair Experiment

An experiment is said to be fair if all possible outcomes are equally likely.

For a fair coin: \[ P(\text{Head}) = P(\text{Tail}) = \frac{1}{2} \]

A fair method ensures no bias and gives equal opportunity to all participants.

Solution Roadmap

Identify number of possible outcomes.
Check whether outcomes are equally likely.
Verify absence of bias.
Conclude fairness based on equal probability.

Solution

Step 1: Identify possible outcomes

When a coin is tossed, there are two possible outcomes:

\[ \text{Head or Tail} \]

Step 2: Assign probabilities

For a fair coin:

\[ P(\text{Head}) = \frac{1}{2}, \quad P(\text{Tail}) = \frac{1}{2} \]

Step 3: Compare chances

Since both outcomes have equal probability, neither outcome is favoured.

Step 4: Application in the game

One team selects "Head" and the other selects "Tail".

Therefore, each team has an equal probability of winning:

\[ P(\text{winning}) = \frac{1}{2} \]

Final Conclusion:

Tossing a coin is a fair method because it gives both teams an equal chance of winning, ensuring a completely unbiased decision.

Exam Significance

Frequently asked as a reason-based conceptual question in CBSE exams.
Tests understanding of equally likely outcomes and fairness.
Foundation for advanced probability problems in higher classes.
Useful in competitive exams like NTSE, CUET, SSC.

Theory: Range of Probability

The probability of any event always lies between 0 and 1 (inclusive).

\[ 0 \leq P(E) \leq 1 \]

\( P(E) = 0 \) → Impossible event
\( 0 < P(E) < 1 \) → Possible event
\( P(E) = 1 \) → Sure event

Solution Roadmap

Convert all values into decimal form.
Check whether each value lies between 0 and 1.
Reject values outside this interval.

(A) \( \frac{2}{3} \)
(B) –1.5
(C) 15%
(D) 0.7

Solution

Step 1: Convert values into decimal form

(A) \[ \frac{2}{3} = 0.666\ldots \]

(C) \[ 15\% = \frac{15}{100} = 0.15 \]

(D) \[ 0.7 = 0.7 \]

Step 2: Check range condition

(A) \(0.666\ldots\) lies between 0 and 1 → Valid
(C) \(0.15\) lies between 0 and 1 → Valid
(D) \(0.7\) lies between 0 and 1 → Valid

(B) \[ -1.5 < 0 \]

Since probability cannot be negative, this value is not valid.

Final Answer: (B) –1.5 cannot be the probability of an event.

Exam Significance

Very common MCQ / 1-mark question in CBSE exams.
Tests understanding of probability bounds.
Useful for eliminating options quickly in competitive exams.
Frequently used in exams like NTSE, SSC, Banking, CUET.

Theory: Complement of an Event

The complement of an event E is the event ‘not E’, denoted by \( E' \).

The sum of probabilities of an event and its complement is always:

\[ P(E) + P(E') = 1 \]

Solution Roadmap

Use complement rule of probability.
Substitute given value.
Simplify step-by-step.

Solution

Step 1: Write the complement formula

\[ P(E) + P(E') = 1 \]

Step 2: Substitute the given value

\[ 0.05 + P(E') = 1 \]

Step 3: Solve for \( P(E') \)

\[ P(E') = 1 - 0.05 \] \[ P(E') = 0.95 \]

Final Answer: Probability of ‘not E’ = 0.95

Exam Significance

Direct application of complement rule, frequently asked in CBSE exams.
Common 1-mark or 2-mark question.
Useful shortcut in complex probability problems.
Important for exams like NTSE, CUET, SSC.

Theory: Impossible and Sure Events

An impossible event has no favourable outcomes: \[ P(\text{impossible event}) = 0 \]
A sure (certain) event includes all possible outcomes: \[ P(\text{sure event}) = 1 \]

Solution Roadmap

Identify total possible outcomes.
Check whether the required outcome exists.
Apply definition of probability.
Classify as impossible or sure event.

an orange-flavoured candy
a lemon-flavoured candy

Solution

Given: The bag contains only lemon-flavoured candies.

(i) Probability of taking out an orange-flavoured candy

Step 1: Identify favourable outcomes

There are no orange-flavoured candies in the bag.

Step 2: Apply probability definition

\[ \begin{aligned} P(\text{orange candy}) &= \frac{\text{Number of favourable outcomes}}{\text{Total outcomes}} \\&= \frac{0}{\text{Total}} \\&= 0 \end{aligned} \]

Conclusion: This is an impossible event.

(ii) Probability of taking out a lemon-flavoured candy

Step 1: Identify favourable outcomes

All candies in the bag are lemon-flavoured.

Step 2: Apply probability definition

\[ P(\text{lemon candy}) = \frac{\text{Total favourable outcomes}}{\text{Total outcomes}} = 1 \]

Conclusion: This is a sure event.

Final Answers:
(i) 0
(ii) 1

Exam Significance

Tests understanding of impossible and sure events.
Frequently appears as a conceptual 1–2 mark question.
Helps build intuition for probability limits.
Important for exams like NTSE, CUET, SSC.

Theory: Complementary Events

If E is an event, then the probability of its complement \( \overline{E} \) is given by:

\[ P(E) + P(\overline{E}) = 1 \]

or,

\[ P(\overline{E}) = 1 - P(E) \]

Solution Roadmap

Identify the given probability.
Recognize the required event as complement.
Apply complement formula.
Compute carefully (avoid arithmetic mistakes).

Solution

Step 1: Define the event

Let E = event that 2 students do not have the same birthday.

\[ P(E) = 0.992 \]

Step 2: Identify required probability

Required: Probability that 2 students have the same birthday.

This is the complement of E:

\[ P(\overline{E}) \]

Step 3: Apply complement formula

\[ P(\overline{E}) = 1 - P(E) \]

Step 4: Substitute value

\[ P(\overline{E}) = 1 - 0.992 \]

Step 5: Calculate

\[ P(\overline{E}) = 0.008 \]

Final Answer: Probability that 2 students have the same birthday = 0.008

Exam Significance

Direct application of complement rule.
Frequently asked in 1–2 mark questions.
Tests both concept clarity and arithmetic accuracy.
Important for exams like NTSE, CUET, SSC.

Theory: Classical Probability

If all outcomes are equally likely, then:

\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Also, for any event E:

\[ P(\text{not }E) = 1 - P(E) \]

Solution Roadmap

Find total number of balls.
Count favourable outcomes for each event.
Apply probability formula.
Use complement rule where needed.

red
not red

Solution

Step 1: Total number of outcomes

\[ \text{Total balls} = 3 + 5 = 8 \]

(i) Probability of drawing a red ball

Step 2: Favourable outcomes

\[ \text{Number of red balls} = 3 \]

Step 3: Apply probability formula

\[ P(\text{red}) = \frac{3}{8} \]

(ii) Probability of drawing a ball that is not red

Method 1: Direct counting

\[ \text{Number of black balls} = 5 \] \[ P(\text{not red}) = \frac{5}{8} \]

Method 2: Using complement rule

\[ \begin{aligned} P(\text{not red}) &= 1 - P(\text{red}) \\&= 1 - \frac{3}{8} \\&= \frac{5}{8} \end{aligned} \]

Final Answers:
(i) \( \frac{3}{8} \)
(ii) \( \frac{5}{8} \)

Exam Significance

Classic probability formula application question.
Tests both direct counting and complement method.
Very common in CBSE board exams (2–3 marks).
Important for competitive exams like NTSE, CUET, SSC.

Theory: Probability with Multiple Outcomes

When outcomes are equally likely:

\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Also, for complement:

\[ P(\overline{E}) = 1 - P(E) \]

Solution Roadmap

Find total number of marbles.
Count favourable outcomes for each case.
Apply probability formula.
Use complement or direct counting.

red
white
not green

Solution

Step 1: Total number of outcomes

\[ \text{Total marbles} = 5 + 8 + 4 = 17 \]

(i) Probability of drawing a red marble

\[ P(\text{red}) = \frac{5}{17} \]

(ii) Probability of drawing a white marble

\[ P(\text{white}) = \frac{8}{17} \]

(iii) Probability of drawing a marble that is not green

Method 1: Direct counting

\[ \text{Not green} = \text{red} + \text{white} = 5 + 8 = 13 \] \[ P(\text{not green}) = \frac{13}{17} \]

Method 2: Complement method

\[ P(\text{green}) = \frac{4}{17} \] \[ P(\text{not green}) = 1 - \frac{4}{17} = \frac{13}{17} \]

Final Answers:
(i) \( \frac{5}{17} \)
(ii) \( \frac{8}{17} \)
(iii) \( \frac{13}{17} \)

Exam Significance

Standard 3-mark probability problem in CBSE exams.
Tests clarity in multi-category outcomes.
Reinforces use of complement method.
Important for exams like NTSE, CUET, SSC.

Theory: Probability with Multiple Categories

If all outcomes are equally likely:

\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

For complement:

\[ P(\text{not }E) = 1 - P(E) \]

Solution Roadmap

Find total number of coins.
Identify favourable outcomes for each case.
Apply probability formula.
Use complement where needed.

is a 50p coin
is not a ₹5 coin

Solution

Step 1: Total number of outcomes

\[ \text{Total coins} = 100 + 50 + 20 + 10 = 180 \]

(i) Probability of drawing a 50p coin

Step 2: Favourable outcomes

\[ \text{Number of 50p coins} = 100 \]

Step 3: Apply formula

\[ P(\text{50p}) = \frac{100}{180} \]

Step 4: Simplify

\[ P(\text{50p}) = \frac{5}{9} \]

(ii) Probability of drawing a coin that is not ₹5

Method 1: Direct counting

\[ \text{Coins not ₹5} = 180 - 10 = 170 \] \[ P(\text{not ₹5}) = \frac{170}{180} = \frac{17}{18} \]

Method 2: Complement method

\[ P(\text{₹5}) = \frac{10}{180} = \frac{1}{18} \] \[ P(\text{not ₹5}) = 1 - \frac{1}{18} = \frac{17}{18} \]

Final Answers:
(i) \( \frac{5}{9} \)
(ii) \( \frac{17}{18} \)

Exam Significance

Classic multi-category probability question (2–3 marks).
Tests both direct counting and complement approach.
Very common in CBSE board exams.
Important for exams like NTSE, CUET, SSC.

Theory: Probability of Selecting an Object

When one object is selected at random from a group of equally likely objects:

\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Solution Roadmap

Identify total number of fish.
Identify favourable outcomes (male fish).
Apply probability formula.

Solution

Step 1: Total number of outcomes

\[ \text{Total fish} = 5 + 8 = 13 \]

Step 2: Number of favourable outcomes

\[ \text{Number of male fish} = 5 \]

Step 3: Apply probability formula

\[ P(\text{male}) = \frac{5}{13} \]

Final Answer: Probability that the fish taken out is a male fish = \(\frac{5}{13}\)

Exam Significance

Direct application of basic probability formula.
Common 1–2 mark question in CBSE exams.
Tests clarity in identifying favourable outcomes.
Important for exams like NTSE, CUET, SSC.

Theory: Equally Likely Outcomes

When all outcomes are equally likely:

\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Solution Roadmap

List all possible outcomes (1 to 8).
Identify favourable outcomes for each case.
Apply probability formula.
Simplify the result.

8
an odd number
a number greater than 2
a number less than 9

Solution

Step 1: Total outcomes

\[ \begin{aligned} S &= \{1,2,3,4,5,6,7,8\}, \\ n(S) &= 8 \end{aligned} \]

(i) Probability of getting 8

\[ \begin{aligned} \text{Favourable outcomes} &= \{8\}, \\ n &= 1 \end{aligned} \] \[ P(8) = \frac{1}{8} \]

(ii) Probability of getting an odd number

\[ \begin{aligned} \text{Odd numbers} &= \{1,3,5,7\}, \\ n &= 4 \end{aligned} \] \[ \begin{aligned} P(\text{odd}) &= \frac{4}{8} \\&= \frac{1}{2} \end{aligned} \]

(iii) Probability of getting a number greater than 2

\[ \begin{aligned} \text{Numbers} &= \{3,4,5,6,7,8\}, \\ n &= 6 \end{aligned} \] \[ \begin{aligned} P(>2) &= \frac{6}{8} \\&= \frac{3}{4} \end{aligned} \]

(iv) Probability of getting a number less than 9

\[ \begin{aligned} \text{Numbers} &= \{1,2,3,4,5,6,7,8\}, \\ n &= 8 \end{aligned} \] \[ P(<9) = \frac{8}{8} = 1 \]

Final Answers:
(i) \( \frac{1}{8} \)
(ii) \( \frac{1}{2} \)
(iii) \( \frac{3}{4} \)
(iv) \( 1 \)

Exam Significance

Classic sample space-based probability question.
Tests ability to list outcomes correctly.
Very common in CBSE board exams (3–4 marks).
Useful for competitive exams like NTSE, CUET, SSC.

Theory Used

When all outcomes of an experiment are equally likely, the probability of an event is

\[ P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \]

For a standard die, the sample space is:

\[ S=\{1,2,3,4,5,6\} \]

Hence, total number of possible outcomes is

\[ n(S)=6 \]

Solution Roadmap

Write the sample space of a die.
Identify the favourable outcomes for each part carefully.
Apply the probability formula step by step.
Simplify each fraction.

Solution

Step 1: Write the sample space

\[ S=\{1,2,3,4,5,6\} \] \[ n(S)=6 \]

(i) Probability of getting a prime number

Prime numbers on a die are:

\[ \{2,3,5\} \]

So, number of favourable outcomes

\[ n(E)=3 \]

Therefore,

\[ P(\text{prime number})=\frac{3}{6}=\frac{1}{2} \]

(ii) Probability of getting a number lying between 2 and 6

Numbers lying between 2 and 6 are:

\[ \{3,4,5\} \]

So, number of favourable outcomes

\[ n(F)=3 \]

Therefore,

\[ P(\text{number between }2\text{ and }6)=\frac{3}{6}=\frac{1}{2} \]

(iii) Probability of getting an odd number

Odd numbers on a die are:

\[ \{1,3,5\} \]

So, number of favourable outcomes

\[ n(G)=3 \]

Therefore,

\[ P(\text{odd number})=\frac{3}{6}=\frac{1}{2} \]

Final Answers

(i) Probability of getting a prime number = \(\frac{1}{2}\)
(ii) Probability of getting a number lying between 2 and 6 = \(\frac{1}{2}\)
(iii) Probability of getting an odd number = \(\frac{1}{2}\)

Why This Solution Matters for Exams

This question strengthens the idea of writing the sample space correctly before solving.
It is important for CBSE Board Exams because such direct probability questions are common as 1-mark and 2-mark questions.
It also helps competitive entrance exam aspirants practise quick identification of prime numbers, odd numbers, and interval-based outcomes.
Students often lose marks by missing favourable outcomes, so the explicit listing method used here is very useful.

Theory: Probability with Playing Cards

A standard deck of playing cards contains:

Total cards = \(52\)
4 suits: hearts, diamonds (red), clubs, spades (black)
Each suit has 13 cards
Face cards = Jack, Queen, King (3 per suit)

Solution Roadmap

Total outcomes = 52
Identify favourable cards carefully
Apply probability formula
Simplify fractions

Solution

Total number of outcomes

\[ n(S)=52 \]

(i) A king of red colour

Red suits: hearts and diamonds → 2 kings

\[ P=\frac{2}{52}=\frac{1}{26} \]

(ii) A face card

Face cards per suit = 3 → total = \(4 \times 3 = 12\)

\[ P=\frac{12}{52}=\frac{3}{13} \]

(iii) A red face card

Red suits = 2, each has 3 face cards → total = 6

\[ P=\frac{6}{52}=\frac{3}{26} \]

(iv) The jack of hearts

Only one such card exists.

\[ P=\frac{1}{52} \]

(v) A spade

Spade suit contains 13 cards.

\[ P=\frac{13}{52}=\frac{1}{4} \]

(vi) The queen of diamonds

Only one such card exists.

\[ P=\frac{1}{52} \]

Final Answers:
(i) \( \frac{1}{26} \)
(ii) \( \frac{3}{13} \)
(iii) \( \frac{3}{26} \)
(iv) \( \frac{1}{52} \)
(v) \( \frac{1}{4} \)
(vi) \( \frac{1}{52} \)

Exam Significance

Tests understanding of card classification.
Very common CBSE board question (2–4 marks).
Helps in solving advanced probability problems.
Important for exams like NTSE, CUET, SSC.

Theory: Probability without Replacement

When an object is drawn and not replaced, the total number of outcomes changes.

Probability formula:

\[ P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Solution Roadmap

Identify total cards.
Count favourable outcomes.
Adjust total when a card is removed.

Solution

Step 1: Total number of cards

\[ n(S)=5 \]

(i) Probability of getting a queen

Only one queen is present.

\[ P(\text{queen})=\frac{1}{5} \]

(ii) Queen is drawn and removed

Remaining cards = 4

(a) Probability of getting an ace

One ace is present among 4 cards.

\[ P(\text{ace})=\frac{1}{4} \]

(b) Probability of getting a queen

Queen has already been removed.

\[ P(\text{queen})=0 \]

Final Answers:
(i) \( \frac{1}{5} \)
(ii) (a) \( \frac{1}{4} \), (b) \( 0 \)

Exam Significance

Tests understanding of without replacement concept.
Common mistake: forgetting total reduces after first draw.
Important for CBSE and competitive exams.

Theory: Probability from Mixed Objects

When objects are mixed randomly, each has equal chance of being selected.

Solution Roadmap

Find total number of pens.
Identify favourable outcomes (good pens).
Apply probability formula.

Solution

Step 1: Total number of pens

\[ \text{Total} = 12 + 132 = 144 \]

Step 2: Favourable outcomes

\[ \text{Good pens} = 132 \]

Step 3: Apply formula

\[ P(\text{good pen})=\frac{132}{144} \]

Step 4: Simplify

\[ P(\text{good pen})=\frac{11}{12} \]

Final Answer: \( \frac{11}{12} \)

Exam Significance

Classic ratio-based probability question.
Tests simplification accuracy.
Frequently asked in CBSE exams (1–2 marks).

Theory: Probability Without Replacement

When an object is drawn and not replaced, the total number of outcomes changes.

Probability formula:

\[ P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Solution Roadmap

Find total bulbs and defective bulbs.
Compute probability using formula.
Update counts after first draw (without replacement).
Recalculate probability for second draw.

Solution

(i) Probability of drawing a defective bulb

Step 1: Total number of bulbs

\[ \text{Total bulbs} = 20 \]

Step 2: Number of defective bulbs

\[ \text{Defective bulbs} = 4 \]

Step 3: Apply probability formula

\[ P(\text{defective})=\frac{4}{20}=\frac{1}{5} \]

(ii) Second draw (first bulb was NOT defective)

Step 1: Update total bulbs

One non-defective bulb is removed.

\[ \text{Remaining bulbs} = 20 - 1 = 19 \]

Step 2: Update non-defective bulbs

\[ \text{Non-defective bulbs initially} = 20 - 4 = 16 \]

One non-defective bulb is removed:

\[ \text{Remaining non-defective bulbs} = 16 - 1 = 15 \]

Step 3: Apply probability formula

\[ P(\text{non-defective})=\frac{15}{19} \]

Final Answers:
(i) \( \frac{1}{5} \)
(ii) \( \frac{15}{19} \)

Exam Significance

Very important question on without replacement.
Tests logical updating of sample space.
Common 2–3 mark question in CBSE exams.
Highly useful for competitive exams like NTSE, CUET, SSC.

Theory: Counting-Based Probability

When outcomes are equally likely:

\[ P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Here, discs are numbered from 1 to 90.

Solution Roadmap

Identify total number of discs.
List/count favourable numbers carefully.
Apply probability formula.

a two-digit number
a perfect square number
a number divisible by 5

Solution

Step 1: Total outcomes

\[ n(S)=90 \]

(i) Probability of a two-digit number

Two-digit numbers range from 10 to 90.

\[ \text{Count} = 90 - 10 + 1 = 81 \] \[ P(\text{two-digit})=\frac{81}{90}=\frac{9}{10} \]

(ii) Probability of a perfect square number

Perfect squares less than or equal to 90:

\[ 1,4,9,16,25,36,49,64,81 \] \[ \text{Count} = 9 \] \[ P(\text{perfect square})=\frac{9}{90}=\frac{1}{10} \]

(iii) Probability of a number divisible by 5

Numbers divisible by 5 between 1 and 90:

\[ 5,10,15,\dots,90 \] \[ \text{Count} = \frac{90}{5} = 18 \] \[ P(\text{divisible by 5})=\frac{18}{90}=\frac{1}{5} \]

Final Answers:
(i) \( \frac{9}{10} \)
(ii) \( \frac{1}{10} \)
(iii) \( \frac{1}{5} \)

Exam Significance

Tests number system concepts inside probability.
Common 3–4 mark question in CBSE exams.
Important for competitive exams (NTSE, CUET, SSC).
Students often lose marks in counting — careful listing is key.

Theory: Probability with Repeated Outcomes

When outcomes repeat, probability depends on how many times a result appears.

\[ P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Solution Roadmap

Identify total faces of the die.
Count how many times each letter appears.
Apply probability formula.

A
D

Solution

Step 1: Total outcomes

\[ n(S)=6 \]

(i) Probability of getting A

Letter A appears 2 times.

\[ P(A)=\frac{2}{6}=\frac{1}{3} \]

(ii) Probability of getting D

Letter D appears 1 time.

\[ P(D)=\frac{1}{6} \]

Final Answers:
(i) \( \frac{1}{3} \)
(ii) \( \frac{1}{6} \)

Exam Significance

Tests concept of repeated outcomes in probability.
Common 1–2 mark question in CBSE exams.
Important for logical counting in competitive exams.

Theory: Geometric Probability

When outcomes depend on area:

\[ P(E)=\frac{\text{Favourable area}}{\text{Total area}} \]

Solution Roadmap

Find area of rectangular region.
Find radius and area of circle.
Apply probability formula.

Solution

Step 1: Area of rectangle

\[ \text{Length} = 3\,m,\quad \text{Breadth} = 2\,m \] \[ \text{Area of rectangle} = 3 \times 2 = 6\,m^2 \]

Step 2: Radius of circle

\[ \text{Diameter} = 1\,m \Rightarrow \text{Radius} = \frac{1}{2} = 0.5\,m \]

Step 3: Area of circle

\[ A_{\text{circle}}=\pi r^2 \] \[ = \pi (0.5)^2 = \pi \times 0.25 = \frac{\pi}{4} \]

Step 4: Apply probability formula

\[ P(E)=\frac{\text{Area of circle}}{\text{Area of rectangle}} \] \[ P(E)=\frac{\frac{\pi}{4}}{6} \]

Step 5: Simplify

\[ P(E)=\frac{\pi}{24} \]

Final Answer: \( \frac{\pi}{24} \)

Exam Significance

Important example of geometric probability.
Tests integration of mensuration + probability.
Common 3–4 mark question in CBSE exams.
Useful for higher-level exams and aptitude tests.

Theory: Complementary Events

If E is an event, then:

\[ P(E)+P(\overline{E})=1 \]

Also,

\[ P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Solution Roadmap

Find total pens and classify into good and defective.
Compute probability of selecting a good pen.
Use complement rule for defective case.

she will buy it
she will not buy it

Solution

Step 1: Total number of pens

\[ n(S)=144 \]

Step 2: Number of good pens

\[ \text{Good pens}=144-20=124 \]

(i) Probability that she will buy the pen (good pen)

\[ P(\text{buy})=\frac{124}{144} \]

Step 3: Simplify

\[ P(\text{buy})=\frac{31}{36} \]

(ii) Probability that she will not buy the pen (defective pen)

Method 1: Complement method

\[ P(\text{not buy})=1-P(\text{buy})=1-\frac{31}{36}=\frac{5}{36} \]

Method 2: Direct method

\[ P(\text{not buy})=\frac{20}{144}=\frac{5}{36} \]

Final Answers:
(i) \( \frac{31}{36} \)
(ii) \( \frac{5}{36} \)

Exam Significance

Tests understanding of real-life probability interpretation.
Important application of complement rule.
Common 2–3 mark question in CBSE exams.
Useful for competitive exams (NTSE, CUET, SSC).

Q22

NUMERIC3 marks

efer to Example 13. (i) Complete the following table:

Event: ‘Sum on 2 dice’	2	3	4	5	6	7	8	9	10	11	12
Probability	\(\dfrac{1}{36}\)	\(\color{orange}\dfrac{2}{36}\)	\(\color{orange}\dfrac{3}{36}\)	\(\color{orange}\dfrac{4}{36}\)	\(\color{orange}\dfrac{5}{36}\)	\(\color{orange}\dfrac{6}{36}\)	\(\dfrac{5}{36}\)	\(\color{orange}\dfrac{4}{36}\)	\(\color{orange}\dfrac{3}{36}\)	\(\color{orange}\dfrac{2}{36}\)	\(\dfrac{1}{36}\)

Theory: Sum of Two Dice

When two dice are thrown, total outcomes are:

\[ 6 \times 6 = 36 \]

Each outcome is equally likely and represented as an ordered pair \((i,j)\).

Solution Roadmap

Total outcomes = 36
List all ordered pairs for each sum
Count favourable outcomes
Apply probability formula

Solution

Total possible outcomes

\[ n(S)=36 \]

Sum = 2

\[ (1,1) \Rightarrow P=\frac{1}{36} \]

Sum = 3

\[ (1,2),(2,1) \Rightarrow P=\frac{2}{36} \]

Sum = 4

\[ (1,3),(2,2),(3,1) \Rightarrow P=\frac{3}{36} \]

Sum = 5

\[ (1,4),(2,3),(3,2),(4,1) \Rightarrow P=\frac{4}{36} \]

Sum = 6

\[ (1,5),(2,4),(3,3),(4,2),(5,1) \Rightarrow P=\frac{5}{36} \]

Sum = 7

\[ (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \Rightarrow P=\frac{6}{36} \]

Sum = 8

\[ (2,6),(3,5),(4,4),(5,3),(6,2) \Rightarrow P=\frac{5}{36} \]

Sum = 9

\[ (3,6),(4,5),(5,4),(6,3) \Rightarrow P=\frac{4}{36} \]

Sum = 10

\[ (4,6),(5,5),(6,4) \Rightarrow P=\frac{3}{36} \]

Sum = 11

\[ (5,6),(6,5) \Rightarrow P=\frac{2}{36} \]

Sum = 12

\[ (6,6) \Rightarrow P=\frac{1}{36} \]

Final Observation: Probabilities increase from sum 2 to 7 and then decrease symmetrically.

Exam Significance

Very important for sample space understanding.
Frequently used in higher probability problems.
Common base for conditional probability and games.
Asked in CBSE and competitive exams.

Theory: Complement Rule in Probability

For any event E:

\[ P(E) + P(\overline{E}) = 1 \]

This is useful when “at least once” type questions are asked.

Solution Roadmap

Total outcomes for two dice = 36
Find outcomes where 5 appears
Use complement for “not coming”

Solution

Step 1: Total outcomes

\[ n(S)=6 \times 6 = 36 \]

Step 2: Count outcomes where 5 appears

5 can appear in:

First throw: (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) → 6 outcomes
Second throw: (1,5),(2,5),(3,5),(4,5),(6,5) → 5 outcomes

Total outcomes where 5 appears = \(6 + 5 = 11\)

(i) Probability that 5 does NOT come at all

Using complement:

\[ P(\text{no 5}) = 1 - P(\text{at least one 5}) \] \[ = 1 - \frac{11}{36} = \frac{25}{36} \]

(ii) Probability that 5 comes at least once

\[ P(\text{at least one 5}) = \frac{11}{36} \]

Final Answers:
(i) \( \frac{25}{36} \)
(ii) \( \frac{11}{36} \)

Shortcut: Instead of counting 11 outcomes, you can directly compute:
Probability of NOT getting 5 in one throw = \( \frac{5}{6} \)
For two throws: \[ P(\text{no 5})=\left(\frac{5}{6}\right)^2=\frac{25}{36} \]

Exam Significance

Very important for “at least once” probability questions.
Tests understanding of complement rule.
Common 3-mark CBSE question.
Highly useful in competitive exams.

Theory: Equally Likely Outcomes

Probability is based on equally likely outcomes in the sample space. Events may consist of one or more outcomes.

Solution Roadmap

Write correct sample space.
Check if outcomes are equally likely.
Count outcomes inside each event carefully.

Solution

(i) Statement about two coins

The argument is not correct.

Correct sample space:

\[ S=\{HH, HT, TH, TT\} \]

These are 4 equally likely outcomes.

Event breakdown:

Two heads → \(HH\) → 1 outcome
Two tails → \(TT\) → 1 outcome
One of each → \(HT, TH\) → 2 outcomes

Probabilities:

\[ P(\text{two heads})=\frac{1}{4} \] \[ P(\text{two tails})=\frac{1}{4} \] \[ P(\text{one of each})=\frac{2}{4}=\frac{1}{2} \]

These are not equal, so the claim of \( \frac{1}{3} \) each is incorrect.

(ii) Statement about a die

The argument is correct.

Sample space:

\[ S=\{1,2,3,4,5,6\} \]

Odd numbers:

\[ \{1,3,5\} \Rightarrow 3 \text{ outcomes} \]

Even numbers:

\[ \{2,4,6\} \Rightarrow 3 \text{ outcomes} \]

Probability:

\[ P(\text{odd})=\frac{3}{6}=\frac{1}{2} \]

Final Conclusion:
(i) Incorrect
(ii) Correct

Exam Significance

Tests conceptual clarity of sample space vs event.
Very common reasoning-based CBSE question.
Important for avoiding logical mistakes in probability.
Highly relevant for competitive exams.

Theory: Probability with Playing Cards

A standard deck of playing cards contains:

Total cards = \(52\)
4 suits: hearts, diamonds (red), clubs, spades (black)
Each suit has 13 cards
Face cards = Jack, Queen, King (3 per suit)

Solution Roadmap

Total outcomes = 52
Identify favourable cards carefully
Apply probability formula
Simplify fractions

Solution

Total number of outcomes

\[ n(S)=52 \]

(i) A king of red colour

Red suits: hearts and diamonds → 2 kings

\[ P=\frac{2}{52}=\frac{1}{26} \]

(ii) A face card

Face cards per suit = 3 → total = \(4 \times 3 = 12\)

\[ P=\frac{12}{52}=\frac{3}{13} \]

(iii) A red face card

Red suits = 2, each has 3 face cards → total = 6

\[ P=\frac{6}{52}=\frac{3}{26} \]

(iv) The jack of hearts

Only one such card exists.

\[ P=\frac{1}{52} \]

(v) A spade

Spade suit contains 13 cards.

\[ P=\frac{13}{52}=\frac{1}{4} \]

(vi) The queen of diamonds

Only one such card exists.

\[ P=\frac{1}{52} \]

Final Answers:
(i) \( \frac{1}{26} \)
(ii) \( \frac{3}{13} \)
(iii) \( \frac{3}{26} \)
(iv) \( \frac{1}{52} \)
(v) \( \frac{1}{4} \)
(vi) \( \frac{1}{52} \)

Exam Significance

Tests understanding of card classification.
Very common CBSE board question (2–4 marks).
Helps in solving advanced probability problems.
Important for exams like NTSE, CUET, SSC.

Symbol / Term	Meaning	Example
P(E)	Probability of event E occurring	P(Head) = 1/2
P(Ē)	Probability of event E NOT occurring (complement)	P(not Head) = 1/2
S	Sample space — set of all possible outcomes	S = {H, T} for a coin
n(S)	Number of elements in sample space	n(S) = 6 for a die
n(E)	Number of favourable outcomes for event E	n(prime on die) = 3
A ∪ B	Union — A or B (or both) occur	King or Heart
A ∩ B	Intersection — both A and B occur	King AND Heart = King of Hearts
A − B	A but not B	King but not Heart (3 cards)
Mutually Exclusive	Events that cannot occur simultaneously	Getting Head and Tail in 1 toss
Exhaustive Events	Events that together cover the entire sample space	{Head, Tail} for a coin toss
Equally Likely	Each outcome has an equal chance of occurring	Each side of a fair die
Empirical P	Based on actual experiment: freq/total trials	Getting 3 heads in 10 tosses → P = 3/10
Theoretical P	Based on logic/formula, ideal conditions	P(Head) = 1/2 by symmetry

Chapter 14 — Probability

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Probability
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