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10 Apr 2023, 10:46
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Difficulty:
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Question Stats:
35% (02:49) correct
65%
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wrong
based on 81
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A test contains 3 sections each containing 4 questions. A student needs to solve a total of 8 questions, ensuring he/she picks at least 2 questions from each section. If a student follows the rules of the test, what is the probability that he/she will pick all 4 questions of a section?
(A) 3/11
(B) 3/10
(C) 3/7
(D) 1/2
(E) 3/4
(A) 3/11
(B) 3/10
(C) 3/7
(D) 1/2
(E) 3/4
10 Apr 2023, 11:21
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ChandlerBong
Given :
- Test contains 3 sections each containing 4 questions
- A student needs to solve a total of 8 questions
Condition: He/she picks at least 2 questions from each section
There are two ways the student can select the questions.
Case 1: 4 questions from any one section & 2 questions from the remaining two sections
Number of ways the student can select one section from which he/she will attempt all four questions = \(^3C_1\)
Number of ways of selecting all four questions from that section = \(^4C_4\)
Number of ways of selecting two questions from the available four questions from the remaining two sections = \(^4C_2*^4C_2\)
Total number of ways = \(^3C_1\)*\(^4C_4\)*\(^4C_2*^4C_2\) =
= 3 * 1 * 6 * 6 = 108
Case 2: 3 questions from any two section & 2 questions from the remaining one section
Number of ways the student can select two sections from which he/she will attempt three out of available four questions = \(^3C_2\)
Number of ways of selecting three questions of the available four questions from that section = \(^4C_3\) * \(^4C_3\)
Number of ways of selecting two questions from the available four questions from the remaining one sections = \(^4C_2\)
Total number of ways = \(^3C_2\)*\(^4C_3\)*\(^4C_3*^4C_2\) = 288
Question : what is the probability that he/she will pick all 4 questions of a section
The favorable case is the result of "Case 1" because in the case, we are considering the scenario in which the student picks up four questions of a given section.
Total cases is represented by the sum of the result of "Case 1" and "Case 2"
Required Probability = \(\frac{108 }{ 108 + 288} = \frac{108 }{ 396} = \frac{3}{11}\)
Option A
23 Apr 2025, 04:23
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