Question 1

Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P(F|E)

Accepted Answer

Given: P(E) = 0.6, P(F) = 0.3, P(E ∩ F) = 0.2

We know that:
P(E|F) = P(E ∩ F) / P(F) = 0.2 / 0.3 = 2/3 ≈ 0.6667

P(F|E) = P(E ∩ F) / P(E) = 0.2 / 0.6 = 1/3 ≈ 0.3333 — Using the definition of conditional probability:
P(A|B) = P(A ∩ B) / P(B)
Calculate P(E|F) and P(F|E) by substituting the given values.

Question 2

Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

Accepted Answer

Given: P(B) = 0.5, P(A ∩ B) = 0.32

Using conditional probability formula:
P(A|B) = P(A ∩ B) / P(B) = 0.32 / 0.5 = 0.64 — Apply the formula P(A|B) = P(A ∩ B) / P(B) with the given values.

Question 3

If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)

Accepted Answer

Given: P(A) = 0.8, P(B) = 0.5, P(B|A) = 0.4

(i) P(A ∩ B) = P(A) × P(B|A) = 0.8 × 0.4 = 0.32

(ii) P(A|B) = P(A ∩ B) / P(B) = 0.32 / 0.5 = 0.64

(iii) P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.8 + 0.5 - 0.32 = 0.98 — Use the formulas:
P(A ∩ B) = P(A) × P(B|A)
P(A|B) = P(A ∩ B) / P(B)
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Substitute the given values to find the answers.

Question 4

Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) = 6/7

Accepted Answer

Given:
2P(A) = P(B) = 5/13
So, P(B) = 5/13
Then, P(A) = (5/13) / 2 = 5/26

P(A|B) = 6/7

We know:
P(A|B) = P(A ∩ B) / P(B) => P(A ∩ B) = P(A|B) × P(B) = (6/7) × (5/13) = 30/91

Now,
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = (5/26) + (5/13) - (30/91)

Calculate:
5/26 = 35/182
5/13 = 70/182
30/91 = 60/182

So,
P(A ∪ B) = (35/182) + (70/182) - (60/182) = (105 - 60)/182 = 45/182 ≈ 0.2473 — First find P(A) from the given relation 2P(A) = P(B).
Use P(A|B) to find P(A ∩ B).
Then use the formula for union of two events.
Convert all fractions to a common denominator to simplify addition and subtraction.

Question 5

If P(A) = 11/11, P(B) = 11/11 and P(A ∪ B) = 11/11, find (i) P(A∩B) (ii) P(A|B) (iii) P(B|A)

Accepted Answer

Given:
P(A) = 11/11 = 1
P(B) = 11/11 = 1
P(A ∪ B) = 11/11 = 1

(i) Using formula:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
1 = 1 + 1 - P(A ∩ B)
=> P(A ∩ B) = 1 + 1 - 1 = 1

(ii) P(A|B) = P(A ∩ B) / P(B) = 1 / 1 = 1

(iii) P(B|A) = P(A ∩ B) / P(A) = 1 / 1 = 1 — Since all probabilities are 1, events A and B are certain.
Use the union formula to find intersection.
Then use conditional probability formulas.

Question 6

A coin is tossed three times, where
(i) E : head on third toss , F : heads on first two tosses
(ii) E : at least two heads , F : at most two heads
(iii) E : at most two tails , F : at least one tail

Accepted Answer

For each part, find P(E|F) and P(F|E) using the definition of conditional probability.

(i) Sample space for three tosses has 8 outcomes.
E: head on third toss = {HHT, HHH, THT, THH}
F: heads on first two tosses = {HHT, HHH}

P(E|F) = P(E ∩ F)/P(F) = P({HHT, HHH})/P({HHT, HHH}) = 1
P(F|E) = P(E ∩ F)/P(E) = P({HHT, HHH})/P({HHT, HHH, THT, THH}) = 2/4 = 1/2

(ii) E: at least two heads = {HHT, HTH, THH, HHH}
F: at most two heads = all except HHH, so {all except HHH}

P(E|F) = P(E ∩ F)/P(F) = P({HHT, HTH, THH})/P({all except HHH}) = 3/7
P(F|E) = P(E ∩ F)/P(E) = P({HHT, HTH, THH})/P({HHT, HTH, THH, HHH}) = 3/4

(iii) E: at most two tails = all outcomes except TTT
F: at least one tail = all except HHH

P(E|F) = P(E ∩ F)/P(F) = P({all except TTT and HHH})/P({all except HHH}) = 6/7
P(F|E) = P(E ∩ F)/P(E) = P({all except TTT and HHH})/P({all except TTT}) = 6/7 — Using the sample space of tossing a coin three times (8 outcomes), events E and F are identified. Then, conditional probabilities are computed using P(E|F) = P(E ∩ F)/P(F) and P(F|E) = P(E ∩ F)/P(E). The intersection and probabilities are calculated by counting favorable outcomes over total outcomes.

Question 7

Two coins are tossed once, where
(i) E : tail appears on one coin, F : one coin shows head
(ii) E : no tail appears, F : no head appears

Accepted Answer

Sample space for tossing two coins: {HH, HT, TH, TT}

(i) E: tail appears on one coin = {HT, TH}
F: one coin shows head = {HT, TH}

P(E|F) = P(E ∩ F)/P(F) = P({HT, TH})/P({HT, TH}) = 1
P(F|E) = P(E ∩ F)/P(E) = P({HT, TH})/P({HT, TH}) = 1

(ii) E: no tail appears = {HH}
F: no head appears = {TT}

P(E|F) = P(E ∩ F)/P(F) = P(∅)/P({TT}) = 0
P(F|E) = P(E ∩ F)/P(E) = P(∅)/P({HH}) = 0 — The sample space has 4 outcomes. Events E and F are identified and their intersections found. Conditional probabilities are computed using the formula P(E|F) = P(E ∩ F)/P(F). Since E and F are disjoint in part (ii), conditional probabilities are zero.

Question 8

A die is thrown three times,
E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses

Accepted Answer

Sample space size = 6^3 = 216

E: 4 appears on third toss = outcomes where third toss is 4 = 6 × 6 × 1 = 36 outcomes
F: 6 and 5 appear on first two tosses respectively = outcomes where first toss = 6, second toss = 5, third toss any = 1 × 1 × 6 = 6 outcomes

E ∩ F: first toss = 6, second toss = 5, third toss = 4 → only 1 outcome

P(E) = 36/216 = 1/6
P(F) = 6/216 = 1/36
P(E ∩ F) = 1/216

P(E|F) = P(E ∩ F)/P(F) = (1/216)/(1/36) = 1/6
P(F|E) = P(E ∩ F)/P(E) = (1/216)/(1/6) = 1/36 — Calculate the number of favorable outcomes for E, F, and their intersection. Use the total sample space of 216. Then apply conditional probability formula.

Question 9

Mother, father and son line up at random for a family picture
E : son on one end, F : father in middle

Accepted Answer

Total number of ways to arrange mother (M), father (F), son (S) = 3! = 6

E: son on one end = son at first or last position
Number of arrangements with son at first position: arrange M and F in 2! = 2 ways
Number of arrangements with son at last position: similarly 2 ways
Total favorable for E = 4

F: father in middle = father fixed in middle position
Arrange M and S at ends: 2! = 2 ways

E ∩ F: son on one end and father in middle
Father fixed in middle, son at first or last position → 2 ways

P(E) = 4/6 = 2/3
P(F) = 2/6 = 1/3
P(E ∩ F) = 2/6 = 1/3

P(E|F) = P(E ∩ F)/P(F) = (1/3)/(1/3) = 1
P(F|E) = P(E ∩ F)/P(E) = (1/3)/(2/3) = 1/2 — Calculate total permutations of 3 people. Count favorable outcomes for E, F and their intersection. Then use conditional probability formula.

Question 10

A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Accepted Answer

(a) Given black die = 5 Possible outcomes for red die = {1,2,3,4,5,6} Sum > 9 means sum ≥ 10 Sum = 5 + red die number > 9 ⇒ red die number > 4 So red die must be 5 or 6 Number of favorable outcomes = 2 Total possible outcomes given black die = 6 P(sum > 9 | black die = 5) = 2/6 = 1/3 (b) Given red die < 4 ⇒ red die = {1,2,3} Sum = 8 ⇒ black die = 8 - red die Possible black die values: If red die = 1 ⇒ black die = 7 (not possible) If red die = 2 ⇒ black die = 6 (possible) If red die = 3 ⇒ black die = 5 (possible) Favorable outcomes = 2 Total possible outcomes given red die < 4 = 3 P(sum = 8 | red die < 4) = 2/3 — For (a), fix black die at 5 and count red die outcomes giving sum > 9. For (b), fix red die < 4 and find black die values making sum 8. Use conditional probability formula.

Question 11

A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}
Find
(i) P(E|F) and P(F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F)|G) and P((E ∩ F)|G)

Accepted Answer

Sample space S = {1,2,3,4,5,6}

E = {1,3,5}, F = {2,3}, G = {2,3,4,5}

(i) P(E) = 3/6 = 1/2, P(F) = 2/6 = 1/3
E ∩ F = {3}
P(E ∩ F) = 1/6

P(E|F) = P(E ∩ F)/P(F) = (1/6)/(1/3) = 1/2
P(F|E) = P(E ∩ F)/P(E) = (1/6)/(1/2) = 1/3

(ii) P(G) = 4/6 = 2/3
E ∩ G = {3,5} ∩ {2,3,4,5} = {3,5}
P(E ∩ G) = 2/6 = 1/3

P(E|G) = P(E ∩ G)/P(G) = (1/3)/(2/3) = 1/2
P(G|E) = P(E ∩ G)/P(E) = (1/3)/(1/2) = 2/3

(iii) E ∪ F = {1,2,3,5}
(E ∪ F) ∩ G = {2,3,5}
P((E ∪ F) ∩ G) = 3/6 = 1/2

E ∩ F = {3}
(E ∩ F) ∩ G = {3}
P((E ∩ F) ∩ G) = 1/6

P((E ∪ F)|G) = P((E ∪ F) ∩ G)/P(G) = (1/2)/(2/3) = 3/4
P((E ∩ F)|G) = P((E ∩ F) ∩ G)/P(G) = (1/6)/(2/3) = 1/4 — Calculate probabilities of events and their intersections. Use conditional probability formula P(A|B) = P(A ∩ B)/P(B) for each part.

Question 12

What is the probability of getting Heads when a fair coin is tossed once?

Accepted Answer

0.5 — A fair coin has two equally likely outcomes: Heads and Tails. Probability of Heads = Number of favorable outcomes / Total outcomes = 1/2 = 0.5.

Question 13

A die is rolled once. What is the probability of getting a number greater than 4?

Accepted Answer

\frac{1}{3} — Numbers greater than 4 on a die are 5 and 6, so favorable outcomes = 2. Total outcomes = 6. Probability = 2/6 = 1/3.

Question 14

Which of the following is NOT a random experiment?

Accepted Answer

Measuring the temperature at noon — A random experiment is one where the outcome cannot be predicted with certainty. Measuring temperature at noon is deterministic and predictable, not random.

Question 15

Identify the sample space when a card is drawn from a standard deck of 52 cards.

Accepted Answer

The sample space S consists of all 52 cards in the deck, i.e., S = {all 52 distinct cards including 13 cards each of Spades, Hearts, Diamonds, and Clubs}. — Sample space is the set of all possible outcomes. Since a standard deck has 52 distinct cards, the sample space contains all these cards.

Probability

Probability — Study Notes

Introduction

Random Experiment, Sample Space and Events

Classical Definition of Probability

Practice Questions — Probability

All 7 Chapters in Mathematics Part-II