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Maru [420]
3 years ago
5

An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm e

xam. After all the students had turned in their exam pa- pers, the instructor randomly ordered them before grading. Consider the first 15 graded exam papers. (a) Find the probability that exactly 10 of these are from the second section. Find the probability that at least 10 of these are from the second section. Find the probability that at least 10 of these are from the same section?
Mathematics
1 answer:
uranmaximum [27]3 years ago
7 0

Answer:

<em>The answers are for option (a) 0.2070  (b)0.3798  (c) 0.3938 </em>

Step-by-step explanation:

<em>Given:</em>

<em>Here Section 1 students = 20 </em>

<em> Section 2 students = 30 </em>

<em> Here there are 15 graded exam papers. </em>

<em> (a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070 </em>

<em> (b) Here if x is the number of students copies of section 2 out of 15 exam papers. </em>

<em>  here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15 </em>

<em>Then, </em>

<em> Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798 </em>

<em> (c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1) </em>

<em> so, </em>

<em> Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938 </em>

<em> Note : Here the given distribution is Hyper-geometric distribution </em>

<em> where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>

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