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ECET 2026 Preparation

Day 65 – Probability – Basics, Permutations & Combinations, and Bayes Theorem (ECET 2026)

Concept Notes (Deep Explanation + Examples)

🌟 1️⃣ Introduction to Probability

Probability measures how likely an event is to occur.

If you toss a coin once, possible outcomes = {H, T}.
Favorable outcomes for Head = 1.
Hence

P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}

So,

P(\text{Head}) = \frac{1}{2}

👉 Range: 0 ≤ P(E) ≤ 1

  • P(E)=0 → impossible
  • P(E)=1 → certain

🎲 2️⃣ Types of Probability

  1. Theoretical: Based on mathematical reasoning.
  2. Experimental: Based on actual trials.
  3. Conditional: When event B has already happened:

P(A|B) = \frac{P(A \cap B)}{P(B)}

🧩 Example:
A bag has 3 red & 2 blue balls. One ball drawn → red. What’s the chance of blue next (without replacement)?

P(\text{Blue|Red}) = \frac{2}{4} = \frac{1}{2}


🔢 3️⃣ Permutations and Combinations

Permutation (Arrangement – Order Matters):

{}^nP_r = \frac{n!}{(n - r)!}

🧮 Example:
How many 3-digit numbers can be formed from 1, 2, 3, 4 without repetition?

{}^4P_3 = \frac{4!}{(4 - 3)!} = \frac{24}{1} = 24

So 24 numbers possible.


Combination (Selection – Order Does Not Matter):

{}^nC_r = \frac{n!}{r!(n - r)!}

🧮 Example:
Selecting 2 students from 4:

{}^4C_2 = \frac{4!}{2!(4 - 2)!} = \frac{24}{4} = 6

Hence 6 ways.


🧮 4️⃣ Relation Between P and C

{}^nP_r = {}^nC_r \times r!


🌐 5️⃣ Law of Total Probability

If B₁, B₂, … Bₙ are mutually exclusive and exhaustive events of S, then

P(A) = P(B_1)P(A|B_1) + P(B_2)P(A|B_2) + \dots + P(B_n)P(A|B_n)


🧭 6️⃣ Bayes Theorem (Proper LaTeX Example)

Used to update probabilities when new information is known.

P(B_i|A) = \frac{P(B_i)P(A|B_i)}{\sum_{j=1}^{n} P(B_j)P(A|B_j)}


Example:

A factory has 3 machines A, B, C producing 20%, 30%, 50% of items respectively.
Defective rates are 2%, 3%, and 4%.
Find the probability that a defective item came from C.

Given:
P(A)=0.2,\quad P(B)=0.3,\quad P(C)=0.5

P(D|A)=0.02,\quad P(D|B)=0.03,\quad P(D|C)=0.04

Using Bayes Theorem:

P(C|D)=\frac{P(C)P(D|C)}{P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)}

Substitute values:

P(C|D)=\frac{0.5\times0.04}{(0.2\times0.02)+(0.3\times0.03)+(0.5\times0.04)}

Simplify:

P(C|D)=\frac{0.02}{0.004+0.009+0.02}=\frac{0.02}{0.033}

Result:

P(C|D)=0.606\approx60.6%

✅ Hence, there’s a 60% chance that a defective item came from Machine C.


🧠 7️⃣ Real-World Engineering Applications

  • Computer Science: AI classification, network failure probability
  • Physics: Random motion of particles
  • Chemistry: Molecular collision rates
  • Electronics: Noise analysis in semiconductors
  • Data Analytics: Spam detection, prediction models

🧩 Diagram (description): Three machines A, B, C → each feeds items → some reach a “defective” bin → arrows show reverse probability flow from defect back to machine (C|D).


⚙️ Formulas (Plain LaTeX)

P(E)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}
{}^nP_r=\frac{n!}{(n-r)!}
{}^nC_r=\frac{n!}{r!(n-r)!}
{}^nP_r={}^nC_r\times r!
P(A|B)=\frac{P(A\cap B)}{P(B)}
P(A\cup B)=P(A)+P(B)-P(A\cap B)
P(A')=1-P(A)
P(A\cap B)=P(A)P(B|A)
P(A)=P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+\dots+P(B_n)P(A|B_n)

P(B_i|A)=\frac{P(B_i)P(A|B_i)}{\sum_{j=1}^{n}P(B_j)P(A|B_j)}


🔟 10 MCQs (ECET + GATE Hybrid)

1️⃣ A coin is tossed twice. Probability of getting exactly one head is
A) 1/4 B) 1/2 C) 3/4 D) 1/3

2️⃣ Number of ways to arrange “MATH” is
A) 12 B) 24 C) 6 D) 18

3️⃣ {}^7C_3 equals
A) 21 B) 35 C) 42 D) 14

4️⃣ Probability of getting sum 7 when two dice are rolled is
A) 1/12 B) 1/8 C) 1/6 D) 1/18

5️⃣ If P(A)=0.4, P(B)=0.5 and A,B independent, then P(A ∩ B)=?
A) 0.2 B) 0.1 C) 0.9 D) 0.5

6️⃣ Number of ways to select 2 persons out of 10 is
A) 90 B) 45 C) 20 D) 100

7️⃣ If P(A)=0.7 and P(B|A)=0.5, then P(A ∩ B)=?
A) 0.3 B) 0.35 C) 0.5 D) 0.2

8️⃣ For mutually exclusive events A and B, P(A ∪ B)=
A) P(A)+P(B) B) P(A)–P(B) C) P(A)×P(B) D) None

9️⃣ Bayes theorem is applicable for
A) Independent events B) Mutually exclusive C) Dependent D) Random

🔟 If 4 coins are tossed, probability of getting 2 heads is
A) 3/8 B) 6/16 C) 3/16 D) 1/2


Answer Key

Q No | Answer
1 | B
2 | B
3 | B
4 | A
5 | A
6 | B
7 | B
8 | A
9 | C
10 | D


🧠 MCQ Explanations

1️⃣ (HH, HT, TH, TT) → 1 head in 2 cases → 2/4 = 1/2.
2️⃣ 4! = 24 arrangements.
3️⃣ {}^7C_3=\frac{7!}{3!4!}=35.
4️⃣ Sum 7 has 6 cases → 6/36 = 1/6.
5️⃣ Independent → P(A\cap B)=0.4\times0.5=0.2.
6️⃣ {}^{10}C_2=\frac{10\times9}{2}=45.
7️⃣ P(A\cap B)=P(A)P(B|A)=0.7\times0.5=0.35.
8️⃣ Mutually exclusive → P(A\cup B)=P(A)+P(B).
9️⃣ Bayes → Dependent events.
10️⃣ {}^4C_2(1/2)^4=6/16=3/8.


🎯 Motivation (ECET 2026 Specific)

Probability appears in ECET every year — in Maths and CSE (randomization, network reliability).
Mastering it gives sure 6–8 marks.
Keep revising 1 formula daily and your rank skyrockets 🚀.


📲 CTA (Fixed)

Join our ECET 2026 CSE WhatsApp Group for daily quizzes & study notes:
👉 https://chat.whatsapp.com/GniYuv3CYVDKjPWEN086X9

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