A bag contains 5 red,4 green and 3 black balls.if three balls are drawan out of it at randam,find the probability of drawing exactly 2 red balls.
a> 7/22 b.10/33 c. 7/12 d. 7/11

• we have to select 2 red balls out of 5 and 1 green from 4 or 2 red balls and 1 black ball
  total combination = 5C2*4C1 + 5C2 *3C1 = 70
  total possibility = 12C3 = 220
  
  probability = 70/220 = 7/22 -> option a
• 9 years agoHelpfull: Yes(29) No(2)
• required probability=(5C2*7C1)/(12C3)
  =7/22
• 9 years agoHelpfull: Yes(17) No(1)
• Out of 5 red balls exactly 2 red balls are drawn. Hence the remaining I ball must drawn from 7 balls(4 green & 3 black).
  
  5C2x7C1/12C3 = 70/220 = 7/22
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• 7/22
  coz 5c2*7c1/12c3
• 9 years agoHelpfull: Yes(2) No(0)
• option a
  s=5+4+3=12
  p(s)=12C3=220
  p(E)=5C2*4C1*3C0+5C2*4C0*3C1
  =40+30=70
  p(E)/p(S)=7/22
• 9 years agoHelpfull: Yes(2) No(0)
• total possible events =12C3=220
  favourable events = 5C2*7C1=70
  probability=70/220
• 9 years agoHelpfull: Yes(1) No(0)
• ((5C2)*(7C1))/12C3 = 7/22
  So option a
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• total number of balls 5+4+3 = 12, out of 12 balls 3 balls are drawn
  so n(s) = 12C3 = 110
  exactly 2 red balls so 5C2 = 10
  out of 3 two are red remaining one is in 4+3 = 7,so 7C1 = 7
  10*7 = 10;
  therefor probablity is 70/110 =7/11
• 9 years agoHelpfull: Yes(0) No(5)
• Since there are 5 red balls and (4+3)other ball,so
  no.of possibilities =5C2*7C1/12C3
  =7/22
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• a>7/22
  5c2*7c1/12c3=7/22
  
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• selecting 3 balls from 12 is 12c3 ways= 220
  select 2 red balls out of 5 and ehither 1 ball from green or 1 ball from black is 5c2*4c1 + 5c2*3c1=70
  so, Total possibility is 70/220= 7/22
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• ((5C2*4C1)+(5C2*3C1))/12C3
  
  (40+30)/220=70/220=7/22
• 9 years agoHelpfull: Yes(0) No(0)
• (5c2*7c1)/12c3=7/22
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• 7/22 = 5C2 * 7C1 / 12C3
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• probability will be=5C2*(4C1+3C1)/12C3=70/220=7/22
  option (a)
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