Latest Elitmus Aptitude Question SOLUTION: There are 10 idols in a box. 5 or less of them are defective. Let the number of defective idols be n. Two idols are drawn simultaneously. What is the value of n? I. The probabilit

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(#M40164746) ELITMUS Logical Reasoning question Keep an EYE

There are 10 idols in a box. 5 or less of them are defective. Let the number of defective idols be n. Two idols are drawn simultaneously. What is the value of n?

I. The probability that both of them are defective is 1/15
II. The probability that one of them is defective and the other is non-defective is 7/15.

Options: Data Sufficiency

Asked In Elitmus Wali Nusker (6 years ago)

Unsolved Read Solution (6)

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Bagwan sohel yusuf  (6 years ago)

only 1st
prob = n*(n-1)/ 10*9 = 1/15
n= 3

Is this solution Helpfull? Yes (5) | No (1)  

Chandan Singha  (6 years ago)

**Both I & II can answer independently**


n: Number of defective
Total Number of ways to select 2 item from 10: 10C2
Number of ways to select 2 defective from n: nC2
Prob: nC2/10C2 = 1/15

Number of ways to select 1 defective & 1 non-defective from n is nC1*(10-n)C1
Prob: nC1*(10-n)C1/10C2 = 7/15

Is this solution Helpfull? Yes (4) | No (2)  

Prashant Kumar  (6 years ago)

nC2/10C2=1/15
=>n(n-1)/2*3=1
=>n^2-3*n-6=0
=>(n-3)(n+2)=0
=>n=3

Is this solution Helpfull? Yes (1) | No (1)  

Badal Gupta  (6 years ago)

n: Number of defective
Total Number of ways to select 2 item from 10: 10C2
Number of ways to select 2 defective from n: nC2
Prob: nC2/10C2 = 1/15
n=3

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  (5 years ago)

Since there is only one unknown, no of defective either of statements determines n
1. P(both defective) = (n/10) * (n-1/9) = 1/15
2. P(one defective) = (n/10)*(9-n)/9 = 7/15

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Pratee Chugh  (4 years ago)

1st can solve for n.
2nd will give either n =3 or n=7.
both will give the same probability.
Hence only a is sufficent.

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