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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