Chapter 13 Class 12 Probability
Serial order wise

Question

Anand, Samanyu and Shah of SHORTCUTS  classes were given a problem in Mathematics  whose respective probabilities of solving it  are 1/2, 1/3  and  1/4. They were asked to solve it  independently.

Based on the above data, answer any four of the  following questions.

 

[Case Based] Anand, Samanyu and Shah of SHORTCUTS classes were given - Case Based Questions (MCQ)

part 2 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability

Question 1

The probability that Anand alone solves it is  _______.

(A) 1/4 

(B) 3/4

(C) 11/24   

(D) 17/24

part 3 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability

part 4 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability

Question 2

The probability that problem is not solved is_______.

(A) 1/4 

(B) 3/4

(C) 0 

(D) 11/24

part 5 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability part 6 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability

Question 3

The probability that the problem is solved is _______.

(A) 1/4 

(B) 3/4

(C) 17/24   

(D) 11/24

part 7 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability

Question 4

The probability that exactly one of them solves it is_______.

(A) 1/4 

(B) 3/4

(C) 17/24   

(D) 11/24

part 8 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability part 9 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability part 10 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability

Question 5

The probability that exactly two of them solves it is_______.

(A) 1/4 

(B) 3/4

(C) 17/24   

(D) 11/24

part 11 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability part 12 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability part 13 - Question 5 - Case Based Questions (MCQ) - Serial order wise - Chapter 13 Class 12 Probability

 

 

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Transcript

Question Anand, Samanyu and Shah of SHORTCUTS classes were given a problem in Mathematics whose respective probabilities of solving it are 1/2, 1/3 and 1/4. They were asked to solve it independently. Based on the above data, answer any four of the following questions. Let, A : The event that Anand solves B : The event that Samanyu solves C : The event that Shah solves Given P(A) = 𝟏/𝟐 P(B) = 𝟏/𝟑 P(C) = 𝟏/𝟒 Question 1 The probability that Anand alone solves it is _______. (A) 1/4 (B) 3/4 (C) 11/24 (D) 17/24 P( A solves alone) = P(A solves) × P(B does not solve) × P(C does not solve) = 1/2×(1−1/3)×(1−1/4) = 1/2×2/3×3/4 = 𝟏/𝟒 So, the correct answer is (a) Question 2 The probability that problem is not solved is_______. (A) 1/4 (B) 3/4 (C) 0 (D) 11/24 P(problem is not solved) = P(A does not solve) × P(B does not solve) × P(C does not solve) = (1−1/2)×(1−1/3)×(1−1/4) = 1/2×2/3×3/4 = 𝟏/𝟒 So, the correct answer is (a) Question 3 The probability that the problem is solved is _______. (A) 1/4 (B) 3/4 (C) 17/24 (D) 11/24 P(problem is solved) = 1− P(problem is not solved) = 1 − 1/4 = 𝟑/𝟒 So, the correct answer is (b) Question 4 The probability that exactly one of them solves it is_______. (A) 1/4 (B) 3/4 (C) 17/24 (D) 11/24 P(exactly one of them solves) = P(A will solve) × P(B will not solve) × P(C will not solve) + P (A will not solve) × P (B will solve) × P (C will not solve) + P (A will not solve)× P (B will not solve) × P (C will solve) = 1/2×(1−1/3)×(1−1/4) +(1−1/2)×1/3×(1−1/4) +(1−1/2)×(1−1/3)×1/4 = 1/2×2/3×3/4 +1/2×1/3×3/4 +1/2×2/3×1/4 = 1/4+1/8+1/12 = 𝟏𝟏/𝟐𝟒 So, the correct answer is (d) Question 5 The probability that exactly two of them solves it is_______. (A) 1/4 (B) 3/4 (C) 17/24 (D) 11/24 P(exactly two of them solves) = P(A will solve) × P(B will solve) × P(C will not solve) + P (A will not solve) × P (B will solve) × P (C will solve) + P (A will solve)× P (B will not solve) × P (C will solve) = 1/2×1/3×(1−1/4) +(1−1/2)×1/3×1/4 +1/2×(1−1/3)×1/4 = 1/2×1/3×3/4 +1/2×1/3×1/4 +1/2×2/3×1/4 = 1/8+1/24+1/12 = 6/24 = 𝟏/𝟒 So, the correct answer is (a)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo