A consignment of 20 picture tubes contains 5 defectives. Two tubes are selected one after the other at random. The probability that both are defective assuming that the first tube is not replaced before drawing
the second, would be:

1. 1/16
2. 1/19
3. 1/4
4. none of these

• Ans. 2. 1/19
  
  Explanation: Probability of drawing a defective tube in first draw is 5/20
  Probability of drawing a defective tube in second draw (without replacing first tube) is 4/19.
  Therefore probability of getting both defective is = (5/20)*(4/19) = 1/19
  
• 12 years agoHelpfull: Yes(29) No(0)
• 5c1/20c1 for drawing defective tube in 1st attempt...
  2nd tube has to be drawn without replacing 1st...so at present nly 19 tubes are there and among them nly 4 r deffective...
  so the prob for getting defective in 2nd case is 4c1/19c1
  
  therefore prob is (5c1/20c1)(4c1/19c1)=1/19
• 12 years agoHelpfull: Yes(4) No(0)
• ans- 1/19
  
  solution= ((5c1*4c1)/(20c1*19c1))= 1/19
• 9 years agoHelpfull: Yes(0) No(0)