✨ Magic Probability Explorer ✨

1 Basic Probability
If \( P(A) = \frac{2}{3} \), \( P(B) = \frac{2}{5} \), \( P(A \cup B) = \frac{1}{3} \) then find \( P(A \cap B) \).
We know the formula: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Plug in the given values: 1/3 = 2/3 + 2/5 - P(A ∩ B)
Calculate 2/3 + 2/5 = 10/15 + 6/15 = 16/15
Now we have: 1/3 = 16/15 - P(A ∩ B)
Rearrange to find P(A ∩ B): P(A ∩ B) = 16/15 - 1/3 = 16/15 - 5/15 = 11/15
Final Answer: \( P(A \cap B) = \frac{11}{15} \)
2 Complementary Probability
A and B are two events such that, \( P(A) = 0.42 \), \( P(B) = 0.48 \), and \( P(A \cap B) = 0.16 \). Find: (i) \( P(\text{not } A) \) (ii) \( P(\text{not } B) \) (iii) \( P(A \text{ or } B) \)
Part (i): \( P(\text{not } A) = 1 - P(A) = 1 - 0.42 = 0.58 \)
Part (ii): \( P(\text{not } B) = 1 - P(B) = 1 - 0.48 = 0.52 \)
Part (iii): Use the union formula: P(A or B) = P(A) + P(B) - P(A ∩ B)
Calculate: 0.42 + 0.48 - 0.16 = 0.74
Final Answers: (i) 0.58 (ii) 0.52 (iii) 0.74
3 Mutually Exclusive Events
If A and B are two mutually exclusive events of a random experiment and \( P(\text{not } A) = 0.45 \), \( P(A \cup B) = 0.65 \), then find \( P(B) \).
First find P(A): P(A) = 1 - P(not A) = 1 - 0.45 = 0.55
For mutually exclusive events: P(A ∪ B) = P(A) + P(B)
Plug in values: 0.65 = 0.55 + P(B)
Solve for P(B): P(B) = 0.65 - 0.55 = 0.10
Final Answer: \( P(B) = 0.10 \)
4 Complementary Events
The probability that at least one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find \( P(\overline{A}) + P(\overline{B}) \).
Given: P(A ∪ B) = 0.6 and P(A ∩ B) = 0.2
Use the union formula: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
So: 0.6 = P(A) + P(B) - 0.2 ⇒ P(A) + P(B) = 0.8
We need to find: P(Ā) + P(B̄) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))
Calculate: 2 - 0.8 = 1.2
Final Answer: \( P(\overline{A}) + P(\overline{B}) = 1.2 \)
5 Neither Event
The probability of happening of an event A is 0.5 and that of B is 0.8. If A and B are mutually exclusive events, then find the probability that neither A nor B happen.
This problem has an inconsistency because the sum of probabilities exceeds 1 for mutually exclusive events.
For mutually exclusive events: P(A ∪ B) = P(A) + P(B) = 0.5 + 0.8 = 1.3
But probability can't exceed 1! There's an inconsistency here.
This suggests the events cannot be mutually exclusive with these probabilities.
Final Answer: Invalid problem - probabilities exceed 1 for mutually exclusive events
6 Dice Probability
Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.
Total possible outcomes when rolling two dice: 6 × 6 = 36
Event A: Even number on first die (2, 4, 6) → 3 choices × 6 for second die = 18 outcomes
Event B: Sum is 8 → (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
Intersection: Even on first die AND sum is 8 → (2,6), (4,4), (6,2) → 3 outcomes
Use the union formula: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Calculate: 18/36 + 5/36 - 3/36 = 20/36 = 5/9
Final Answer: \( \frac{5}{9} \)
7 Card Probability
A box contains cards numbered 3, 5, 7, 9, ... 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.
First identify the sequence: 3, 5, 7, 9, ..., 35, 37 (odd numbers from 3 to 37)
Count the numbers: First term (a) = 3, Last term (l) = 37, Difference (d) = 2
Number of terms: n = (37-3)/2 + 1 = 18
Multiples of 7: 7, 21, 35 → 3 numbers
Prime numbers: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 → 11 numbers
Intersection: Numbers that are both prime and multiples of 7 → only 7 → 1 number
Use the union formula: P(A ∪ B) = 3/18 + 11/18 - 1/18 = 13/18
Final Answer: \( \frac{13}{18} \)
8 Coin Toss Probability
Three unbiased coins are tossed once. Find the probability of getting at most 2 tails or at least 2 heads.
Total possible outcomes when tossing 3 coins: 2 × 2 × 2 = 8
Event A: At most 2 tails → All outcomes except TTT → 7 outcomes
Event B: At least 2 heads → HHT, HTH, THH, HHH → 4 outcomes
Intersection: At most 2 tails AND at least 2 heads → same as Event B → 4 outcomes
Use the union formula: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Calculate: 7/8 + 4/8 - 4/8 = 7/8
Final Answer: \( \frac{7}{8} \)
9 Contract Probability
The probability that a person will get an electrification contract is \( \frac{3}{5} \) and the probability that he will not get plumbing contract is \( \frac{5}{8} \). The probability of getting at least one contract is \( \frac{5}{7} \). What is the probability that he will get both?
Given: P(E) = 3/5, P(not P) = 5/8 ⇒ P(P) = 1 - 5/8 = 3/8
Also given: P(E ∪ P) = 5/7
We need to find P(E ∩ P)
Use the union formula: P(E ∪ P) = P(E) + P(P) - P(E ∩ P)
Plug in values: 5/7 = 3/5 + 3/8 - P(E ∩ P)
Calculate 3/5 + 3/8 = 24/40 + 15/40 = 39/40
Now: 5/7 = 39/40 - P(E ∩ P)
Rearrange: P(E ∩ P) = 39/40 - 5/7 = (273-200)/280 = 73/280
Final Answer: \( \frac{73}{280} \)
10 Town Demographics
In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that 30% of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years?
Total population: 8000
Females: 3000
Over 50 years: 1300
Females over 50: 30% of 3000 = 900
Use the union formula: P(F ∪ O) = P(F) + P(O) - P(F ∩ O)
Calculate: 3000/8000 + 1300/8000 - 900/8000 = 3400/8000 = 17/40
Final Answer: \( \frac{17}{40} \)
11 Coin Toss Combinations
A coin is tossed thrice. Find the probability of getting exactly two heads or at least one tail or two consecutive heads.
Total possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT → 8
Event A: Exactly two heads → HHT, HTH, THH → 3 outcomes
Event B: At least one tail → All except HHH → 7 outcomes
Event C: Two consecutive heads → HHT, THH, HHH → 3 outcomes
We need: P(A ∪ B ∪ C)
Notice that Event B covers almost all cases (7/8), and all A and C cases are included in B
Thus, the union is simply Event B
Final Answer: \( \frac{7}{8} \)
12 Three Events Probability
If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if \( P(A \cap B) = \frac{1}{6} \), \( P(B \cap C) = \frac{1}{4} \), \( P(A \cap C) = \frac{1}{8} \), \( P(A \cup B \cup C) = \frac{9}{10} \), \( P(A \cap B \cap C) = \frac{1}{15} \), then find \( P(A), P(B) \) and \( P(C) \)?
Let P(A) = x, then P(B) = 2x, P(C) = 3x
Use the general union formula for three events:
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C)
Plug in given values: 9/10 = x + 2x + 3x - 1/6 - 1/4 - 1/8 + 1/15
Simplify: 9/10 = 6x - (20/120 + 30/120 + 15/120) + 8/120
Calculate: 9/10 = 6x - (65/120) + 8/120 = 6x - 57/120
Convert to common denominator (120): 108/120 = 6x - 57/120
Rearrange: 6x = 165/120 ⇒ x = 165/720 = 11/48
Thus: P(A) = 11/48, P(B) = 22/48 = 11/24, P(C) = 33/48 = 11/16
Final Answers: \( P(A) = \frac{11}{48} \), \( P(B) = \frac{11}{24} \), \( P(C) = \frac{11}{16} \)
13 Class Roll Numbers
In a class of 35, students are numbered from 1 to 35. The ratio of boys to girls is 4:3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.
Total students: 35
Boys:Girls = 4:3 ⇒ Boys = 20, Girls = 15
Roll numbers: Boys 1-20, Girls 21-35
Event A: Boy with prime roll number
Primes in 1-20: 2,3,5,7,11,13,17,19 → 8 numbers
Event B: Girl with composite roll number
Composites in 21-35: 21,22,24,25,26,27,28,30,32,33,34,35 → 12 numbers
Event C: Even roll number
Evens in 1-35: 2,4,6,...,34 → 17 numbers (2-34, step 2)
Intersections:
A ∩ B: Impossible (mutually exclusive)
A ∩ C: Boy with prime and even → only 2 → 1 number
B ∩ C: Girl with composite and even → 22,24,26,28,30,32,34 → 7 numbers
A ∩ B ∩ C: Impossible
Use inclusion-exclusion principle:
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ C) - P(B ∩ C)
Calculate: 8/35 + 12/35 + 17/35 - 1/35 - 7/35 = 29/35
Final Answer: \( \frac{29}{35} \)