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Nalini Uday Srivastava
Asked: 2022-11-10T15:00:08+05:30 In:

In a class of 10 student, probability of exactly I students passing an examination is directly proportional to `i^(2).` Then answer the following questions:
If a student is selected at random, then the probability that he has passed the examination is
A. `1//7`
B. `11//35`
C. `11//14`
D. None of these

In a class of 10 student, probability of exactly I students passing an examination is directly proportional to `i^(2).` Then answer the following questions:
If a student is selected at random, then the probability that he has passed the examination is
A. `1//7`
B. `11//35`
C. `11//14`
D. None of these
Parabola
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  1. Correct Answer – C
    Let P(i) be the probability that exactly I students are passing an examination. Now given that
    `P(A_(i))=lamda_(i)^(2)(where lamda”is constant”)`
    `impliesunderset(i-1)overset(10)sumP(A_(i))=underset(i-1)overset(10)sum lamdai^(2)=lamda(10xx11xx21)/(6)=lamdaxx386=1`
    `implieslamda =1//358=5//77.`
    Let A reprsent the event that selected students have passed the examination. Therefore,
    `P(A)underset(i-1)overset(10)sumP(A//A_(i))P(A_(i))`
    `=underset(i-1)overset(10)sum(i)/(10)(i^(2))/(358)`
    `=(1)/(3850)underset(i-1)overset(10)sumi^(3`
    `=(10^(2)xx11^(2))/(4xx3850)=11/14`
    Now ` P(A_(i)//A)=((PA//A_(i))P(A_(i)))/(P(A))`
    `=(1/358xx1/10)/(11/14)`
    `=(1)/(11xx55)xx1/5=1/3025`

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Alpa Shenoy
Asked: 2022-11-04T04:31:13+05:30 In:

In a class of 10 student, probability of exactly I students passing an examination is directly proportional to `i^(2).` Then answer the following questions:
If a students selected at random is found to have passed the examination, then the probability that he was the only student who has passed the examination is
A. `1//3025`
B. `1//605`
C. `1//275`
D. `1//121`

In a class of 10 student, probability of exactly I students passing an examination is directly proportional to `i^(2).` Then answer the following questions:
If a students selected at random is found to have passed the examination, then the probability that he was the only student who has passed the examination is
A. `1//3025`
B. `1//605`
C. `1//275`
D. `1//121`
Parabola
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1 Answer

  1. Correct Answer – A
    Let P(i) be the probability that exactly I students are passing an examination. Now given that
    `P(A_(i))=lamda_(i)^(2)(where lamda”is constant”)`
    `impliesunderset(i-1)overset(10)sumP(A_(i))=underset(i-1)overset(10)sum lamdai^(2)=lamda(10xx11xx21)/(6)=lamdaxx386=1`
    `implieslamda =1//358=5//77.`
    Let A reprsent the event that selected students have passed the examination. Therefore,
    `P(A)underset(i-1)overset(10)sumP(A//A_(i))P(A_(i))`
    `=underset(i-1)overset(10)sum(i)/(10)(i^(2))/(358)`
    `=(1)/(3850)underset(i-1)overset(10)sumi^(3`
    `=(10^(2)xx11^(2))/(4xx3850)=11/14`
    Now ` P(A_(i)//A)=((PA//A_(i))P(A_(i)))/(P(A))`
    `=(1/358xx1/10)/(11/14)`
    `=(1)/(11xx55)xx1/5=1/3025`

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Munaf Sood
Asked: 2022-11-02T23:25:46+05:30 In:

In a class of 10 student, probability of exactly I students passing an examination is directly proportional to `i^(2).` Then answer the following questions:
The probability that exactly 5 students passing an examination is
A. `1//11`
B. `5//77`
C. `25//77`
D. `10//77`

In a class of 10 student, probability of exactly I students passing an examination is directly proportional to `i^(2).` Then answer the following questions:
The probability that exactly 5 students passing an examination is
A. `1//11`
B. `5//77`
C. `25//77`
D. `10//77`
Parabola
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1 Answer

  1. Correct Answer – B
    Let P(i) be the probability that exactly I students are passing an examination. Now given that
    `P(A_(i))=lamda_(i)^(2)(where lamda”is constant”)`
    `impliesunderset(i-1)overset(10)sumP(A_(i))=underset(i-1)overset(10)sum lamdai^(2)=lamda(10xx11xx21)/(6)=lamdaxx386=1`
    `implieslamda =1//358=5//77.`
    Let A reprsent the event that selected students have passed the examination. Therefore,
    `P(A)underset(i-1)overset(10)sumP(A//A_(i))P(A_(i))`
    `=underset(i-1)overset(10)sum(i)/(10)(i^(2))/(358)`
    `=(1)/(3850)underset(i-1)overset(10)sumi^(3`
    `=(10^(2)xx11^(2))/(4xx3850)=11/14`
    Now ` P(A_(i)//A)=((PA//A_(i))P(A_(i)))/(P(A))`
    `=(1/358xx1/10)/(11/14)`
    `=(1)/(11xx55)xx1/5=1/3025`

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