A bag contains 5 red & 3 black balls and the second one 4 red and 5 black balls. One of is selected at random and a draw of two balls is made from it. What is the chances that one of them is red and the other is black?

• secet 1 bag from two=1/2
  from 1 st bag=5c1*3c1/8c2=15/28
  from 2nd bag=4c1*5c1/9c2=20/36
  toatl p(s)=1/2(15/28 + 20/36)=.5456
• 12 years agoHelpfull: Yes(10) No(0)
• if 1st bag is selected,probability of getting one red and other black is
  = [(5/8)*(3/7)]+[(3/8)*(5/7)]
  
  similarly if bag 2 is selected, probability of getting one red and other black is = [(4/9)*(5/8)]+[(5/9)*(4/8)]
  
  probability of selecting one bag is 1/2
  
  hence answer = (1/2)[(5/8)*(3/7)]+[(3/8)*(5/7)] + (1/2)[(4/9)*(5/8)]+[(5/9)*(4/8)]
  
  =0.5456
  
• 12 years agoHelpfull: Yes(2) No(0)