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

• only 1st
  prob = n*(n-1)/ 10*9 = 1/15
  n= 3
• 7 years agoHelpfull: Yes(5) No(1)
• **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
• 7 years agoHelpfull: Yes(4) No(2)
• nC2/10C2=1/15
  =>n(n-1)/2*3=1
  =>n^2-3*n-6=0
  =>(n-3)(n+2)=0
  =>n=3
• 7 years agoHelpfull: Yes(1) No(1)
• 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
• 7 years agoHelpfull: Yes(0) No(0)
• 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
• 6 years agoHelpfull: Yes(0) No(0)
• 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.
• 5 years agoHelpfull: Yes(0) No(0)