Episode 8 — Aptitude and Reasoning / 8.15 — Probability
8.15.c Solved Examples -- Probability
Basic Level
Example 1: Single Die
Problem: A fair die is rolled. What is the probability of getting a number greater than 4?
Solution:
S = {1, 2, 3, 4, 5, 6} Total outcomes = 6
E (>4) = {5, 6} Favourable = 2
P(E) = 2/6 = 1/3
Answer: 1/3
Example 2: Single Card
Problem: A card is drawn at random from a standard deck of 52 cards. What is the probability of drawing a face card?
Solution:
Face cards = Jack, Queen, King in each of 4 suits = 12
P(face card) = 12/52 = 3/13
Answer: 3/13
Example 3: Coin Toss
Problem: Three fair coins are tossed. What is the probability of getting exactly 2 heads?
Solution:
Total outcomes = 2^3 = 8
Favourable (exactly 2 heads) = 3C2 = 3
{HHT, HTH, THH}
P = 3/8
Answer: 3/8
Example 4: Complementary Probability
Problem: A bag contains 6 red and 4 blue balls. One ball is drawn. What is the probability that it is NOT red?
Solution:
Total = 10
P(red) = 6/10 = 3/5
P(not red) = 1 - 3/5 = 2/5
Answer: 2/5
Example 5: Simple Addition Rule
Problem: A die is rolled. What is the probability of getting a 2 or a 5?
Solution:
These are mutually exclusive events (cannot get both on one roll).
P(2 or 5) = P(2) + P(5) = 1/6 + 1/6 = 2/6 = 1/3
Answer: 1/3
Intermediate Level
Example 6: Two Dice -- Sum
Problem: Two dice are thrown. What is the probability that the sum is 9?
Solution:
Total outcomes = 36
Favourable outcomes for sum = 9:
(3,6), (4,5), (5,4), (6,3) = 4 outcomes
P(sum = 9) = 4/36 = 1/9
Answer: 1/9
Example 7: Cards -- OR (Non-Exclusive)
Problem: A card is drawn from a deck. What is the probability that it is a Spade or an Ace?
Solution:
P(Spade) = 13/52
P(Ace) = 4/52
P(Spade AND Ace) = 1/52 (Ace of Spades)
P(Spade or Ace) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13
Answer: 4/13
Example 8: Without Replacement
Problem: A bag has 5 white and 3 black balls. Two balls are drawn without replacement. What is the probability that both are white?
Solution:
Method 1 (Sequential):
P(1st white) = 5/8
P(2nd white | 1st white) = 4/7
P(both white) = 5/8 x 4/7 = 20/56 = 5/14
Method 2 (Combinations):
P = 5C2 / 8C2 = 10/28 = 5/14
Answer: 5/14
Example 9: At Least One (Complement)
Problem: A coin is tossed 4 times. What is the probability of getting at least one tail?
Solution:
P(at least 1 tail) = 1 - P(no tail)
= 1 - P(all heads)
= 1 - (1/2)^4
= 1 - 1/16
= 15/16
Answer: 15/16
Example 10: Independent Events
Problem: The probability of A solving a problem is 1/3 and the probability of B solving it is 1/4. If they work independently, what is the probability that the problem is solved?
Solution:
"Problem is solved" means at least one of them solves it.
P(A fails) = 1 - 1/3 = 2/3
P(B fails) = 1 - 1/4 = 3/4
P(both fail) = 2/3 x 3/4 = 6/12 = 1/2
P(problem solved) = 1 - P(both fail) = 1 - 1/2 = 1/2
Answer: 1/2
Example 11: Drawing Balls -- Mixed
Problem: A box contains 4 red, 5 green, and 6 white balls. Three balls are drawn at random. What is the probability that one of each colour is drawn?
Solution:
Total balls = 15
Total ways to draw 3 = 15C3 = 455
Favourable = 4C1 x 5C1 x 6C1 = 4 x 5 x 6 = 120
P = 120/455 = 24/91
Answer: 24/91
Example 12: Two Dice -- Doublet
Problem: Two dice are thrown. What is the probability of getting a doublet (both dice showing the same number)?
Solution:
Total outcomes = 36
Doublets: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6
P(doublet) = 6/36 = 1/6
Answer: 1/6
Example 13: Cards -- Two Draws Without Replacement
Problem: Two cards are drawn from a deck without replacement. What is the probability that the first is a King and the second is a Queen?
Solution:
P(1st King) = 4/52
P(2nd Queen | 1st King) = 4/51
P(King then Queen) = 4/52 x 4/51 = 16/2652 = 4/663
Answer: 4/663
Example 14: Multiple Coins -- Exact Count
Problem: Five coins are tossed. What is the probability of getting exactly 3 heads?
Solution:
Total outcomes = 2^5 = 32
Favourable = 5C3 = 10
P(exactly 3 heads) = 10/32 = 5/16
Answer: 5/16
Example 15: Conditional Probability
Problem: In a class, 60% of students pass in Mathematics and 45% pass in Physics. 30% pass in both. If a student passes in Mathematics, what is the probability that the student also passes in Physics?
Solution:
P(M) = 0.60
P(P) = 0.45
P(M and P) = 0.30
P(P|M) = P(M and P) / P(M)
= 0.30 / 0.60
= 0.50
= 1/2
Answer: 1/2
Advanced Level
Example 16: Balls Without Replacement -- Specific Sequence
Problem: A bag contains 3 red, 4 blue, and 5 green balls. Three balls are drawn one by one without replacement. What is the probability of drawing them in the order Red, Blue, Green?
Solution:
P(1st Red) = 3/12 = 1/4
P(2nd Blue | 1st Red) = 4/11
P(3rd Green | 1st Red, 2nd Blue) = 5/10 = 1/2
P(R, B, G in order) = 1/4 x 4/11 x 1/2
= 4/88
= 1/22
Answer: 1/22
Example 17: Probability Using Permutations
Problem: 4 people are randomly arranged in a line. What is the probability that 2 specific people are adjacent?
Solution:
Total arrangements = 4! = 24
Favourable: Treat the 2 specific people as a unit.
Units: {(AB), C, D} = 3 units
Arrangements = 3! x 2! = 6 x 2 = 12
P = 12/24 = 1/2
Answer: 1/2
Example 18: Birthday Problem
Problem: In a group of 4 people, what is the probability that at least 2 have the same birthday? (Assume 365 days, ignore leap years.)
Solution:
P(at least 2 same) = 1 - P(all different)
P(all different) = (365/365) x (364/365) x (363/365) x (362/365)
= 1 x 364/365 x 363/365 x 362/365
= (365 x 364 x 363 x 362) / 365^4
= 47,831,784,360 / 17,748,900,625
≈ 0.9836
Wait, that's P(all different). Let me compute more carefully.
P(all different) = 365/365 x 364/365 x 363/365 x 362/365
= 1 x 0.99726 x 0.99452 x 0.99178
= 0.98364 (approximately)
P(at least 2 same) = 1 - 0.98364 = 0.01636
≈ 1.64%
Answer: Approximately 1.64% (or about 0.0164)
Example 19: Cards -- Poker Hand
Problem: What is the probability of getting a flush (all 5 cards of the same suit) in a 5-card poker hand?
Solution:
Total 5-card hands = 52C5 = 2,598,960
A flush = all 5 cards from one suit.
Ways for one suit = 13C5 = 1287
There are 4 suits, so: 4 x 1287 = 5,148
But this includes straight flushes and royal flushes.
Straight flushes (including royal) per suit = 10
Total straight flushes = 40
Pure flush = 5,148 - 40 = 5,108
P(flush) = 5,108 / 2,598,960 ≈ 0.00197
If the question counts all flushes (including straight flush):
P = 5,148 / 2,598,960 ≈ 0.00198
Answer: 5,108/2,598,960 (approximately 0.197%)
Example 20: Dependent Events -- Sequential
Problem: A bag contains 10 balls numbered 1 to 10. Three balls are drawn one by one without replacement. What is the probability that the numbers are drawn in ascending order?
Solution:
Any 3 balls chosen can be arranged in 3! = 6 orders.
Exactly 1 of these 6 orders is ascending.
P(ascending order) = 1/6
(This is true regardless of which 3 balls are drawn, by symmetry.)
Answer: 1/6
Example 21: Two Dice -- At Least One Six
Problem: Two dice are thrown. What is the probability of getting at least one six?
Solution:
P(at least one 6) = 1 - P(no six on either die)
P(no 6 on one die) = 5/6
P(no 6 on both dice) = 5/6 x 5/6 = 25/36
P(at least one 6) = 1 - 25/36 = 11/36
Answer: 11/36
Example 22: Odds Problem
Problem: The odds in favour of an event are 3:7. What is the probability of the event?
Solution:
Odds in favour = 3:7 means favourable : unfavourable = 3 : 7
P(E) = 3 / (3 + 7) = 3/10
Answer: 3/10
Example 23: Multiple Independent Events
Problem: A, B, and C independently try to solve a problem. Their probabilities of solving it are 1/2, 1/3, and 1/4 respectively. What is the probability that exactly one of them solves it?
Solution:
P(A solves) = 1/2, P(A fails) = 1/2
P(B solves) = 1/3, P(B fails) = 2/3
P(C solves) = 1/4, P(C fails) = 3/4
Exactly one solves:
Case 1: Only A solves = 1/2 x 2/3 x 3/4 = 6/24 = 1/4
Case 2: Only B solves = 1/2 x 1/3 x 3/4 = 3/24 = 1/8
Case 3: Only C solves = 1/2 x 2/3 x 1/4 = 2/24 = 1/12
P(exactly one) = 1/4 + 1/8 + 1/12
= 6/24 + 3/24 + 2/24
= 11/24
Answer: 11/24
Example 24: Cards -- Multiple Conditions
Problem: Three cards are drawn from a deck of 52 cards. What is the probability that all three are from different suits?
Solution:
Total ways = 52C3 = 22,100
Favourable: Choose 3 suits from 4, then 1 card from each chosen suit.
= 4C3 x 13 x 13 x 13
= 4 x 2,197
= 8,788
P = 8,788 / 22,100 = 2197/5525
Simplify: 2197 = 13^3, 5525 = 13 x 425 = 13 x 5 x 85 = 13 x 5 x 5 x 17
Hmm let me recompute.
52C3 = 52 x 51 x 50 / 6 = 22,100
P = 8,788 / 22,100
Divide both by 4: 2,197 / 5,525
2197 = 13^3
5525 = 5 x 1105 = 5 x 5 x 221 = 25 x 221 = 25 x 13 x 17
P = 13^3 / (25 x 13 x 17) = 13^2 / (25 x 17) = 169/425
Answer: 169/425
Example 25: Expected Value
Problem: A game involves rolling a die. You win Rs. 12 if you roll a 6, Rs. 6 if you roll a 3 or 4, and lose Rs. 6 otherwise. What is the expected gain?
Solution:
Outcomes and probabilities:
Roll 6: win Rs. 12 P = 1/6
Roll 3 or 4: win Rs. 6 P = 2/6
Roll 1, 2, or 5: lose Rs. 6 P = 3/6
E(X) = 12(1/6) + 6(2/6) + (-6)(3/6)
= 12/6 + 12/6 - 18/6
= 2 + 2 - 3
= 1
Answer: Expected gain = Rs. 1
Example 26: Balls -- Conditional
Problem: A box has 5 red and 7 blue balls. Two balls are drawn. Given that the first ball drawn is red, what is the probability that the second ball is also red?
Solution:
This is conditional probability (without replacement).
Given 1st ball is red: remaining = 4 red + 7 blue = 11 balls
P(2nd red | 1st red) = 4/11
Answer: 4/11
Example 27: Repeated Trials (Binomial)
Problem: A biased coin has P(Head) = 2/3. It is tossed 5 times. What is the probability of getting exactly 3 heads?
Solution:
Using binomial formula: P(X=k) = nCk x p^k x q^(n-k)
n = 5, k = 3, p = 2/3, q = 1/3
P(X=3) = 5C3 x (2/3)^3 x (1/3)^2
= 10 x 8/27 x 1/9
= 10 x 8/243
= 80/243
Answer: 80/243
Example 28: Probability in Geometry
Problem: A point is randomly selected inside a rectangle of length 12 and width 8. What is the probability that the point lies within 2 units of the centre?
Solution:
Area of rectangle = 12 x 8 = 96
The region within 2 units of the centre is a circle of radius 2.
Area of circle = pi x 2^2 = 4*pi
P = 4*pi / 96 = pi/24 ≈ 0.1309
Answer: pi/24 (approximately 0.131)
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