A basket contains 6 red balls, 5 blue balls, 4 green balls and 3 white balls. Five balls are to be drawn together at random, then what is the probability that there is at least a ball of each colour?

a. 10/51
b. 5/51
c. 10/17
d. 5/17

• required probability={(6*5*4*3c2)+(6*5*4c2*3)+(6*5c2*4*3)+(6c2+5*4*3)}/18c5
  =(360+540+720+900)/18*17*14*2
  =5/17
• 9 years agoHelpfull: Yes(34) No(2)
• 6c2*5c1*4c1*3c1 + 6c1*5c2*4c1*3c1 + 6c1*5c1*4c2*3c1 + 6c1*5c1*4c1*3c2 =2520
  total 18c5
  ans 25208568= 5/17
• 9 years agoHelpfull: Yes(23) No(0)
• as we have to draw 5 balls in which at leat 1 ball of each colour should be present..so it can be done in four ways...by selecing 2 balls from one colour and rest 3 from remaining 3 colours...so required probability=
  {(6c2*5c1*4c1*3c1)+(6c1*5c2*4c1*3c1)+(6c1*5c1*4c2*3c1)+(6c1*5c1*4c1*3c2)}/18c5
  =5/17
• 9 years agoHelpfull: Yes(8) No(0)
• there are total 5 balls selected and we have to select at least one from each color, that leaves that the remaining ball can be of either of 4 colors
  there will be 4 cases
  case 1 : last ball is red
  6c2*5c1*4c1*3c1
  case 2 : blue
  6c1*5c2*4c1*3c1
  and this will go like this for rest 2 cases
  
  now we can select 5 balls from 18 in 18c5 ways
  so the required probability will be
  (6c2*5c1*4c1*3c1 + 6c1*5c2*4c1*3c1 + 6c1*5c1*4c2*3c1 + 6c1*5c1*4c1*3c2)/18c5
  = 5/17
  
• 9 years agoHelpfull: Yes(5) No(0)
• total reqiured possible ways are (6*5*4*3c2)+(6*5*4c2*3)+(6*5c2*4*3)+(6c2*5*4*3)===>>>360+540+720+900=2520
  total ways of selecting 5 balls=18c5=8568
  so probability=2520/8568=5/17
• 9 years agoHelpfull: Yes(3) No(0)
• ANSWER ==> 5/17
  P(E)= {(6C1*5C1*4C1*3C2)+(6C1*5C1*4C2*3C1)+(6C1*5C2*4C1*3C1)+(6C2*5C1*4C1*3C1)}/18C5 = 2520/8568 = 5/17
• 9 years agoHelpfull: Yes(3) No(0)
• p(e)=(3c2*6c1*5c1*4c1)+(3c1*6c2*5c1*4c1)+(3c1*6c1*5c2*4c1)+(3c1*6c1*5c1*4c2)/(18c5)=10/17 ans....
• 9 years agoHelpfull: Yes(3) No(0)
• probability=[(6C2*5*4*3)+(6*5C2*4*3)+(6*5*4C2*3)+(6*5*4*3C2)]/18C5
  on solving =5/17
• 9 years agoHelpfull: Yes(2) No(0)
• probability=((6c2*5c1*4c1*3c1)+(6c1*5c2*4c1*3c1)+(6c1*5c1*4c2*3c1)+(6c1*5c1*4c1*3c2))/18c5
  =5/17
  
• 9 years agoHelpfull: Yes(1) No(0)
• (6c2*5c1*4c1*3c1+6c1*5c2*4c1*3c1+6c1*5c2*4c1*3c1+6c2*5c1*4c1*3c1)/18c5=5/17
• 9 years agoHelpfull: Yes(1) No(0)
• Ans=c
  n(s)=18C5=8568
  n(a)=6c1*5C1*4c1*3c1*14c1=5040
  p9a0=5040/8658
• 9 years agoHelpfull: Yes(1) No(1)
• D
  taking 4 cases with 2 balls frm red,blue,green,white resp. in each case nd rest others balls of other clrs
• 9 years agoHelpfull: Yes(0) No(0)
• (6c1*5c1*4c1*3c1)(14c1)/18c5=10/17
• 9 years agoHelpfull: Yes(0) No(5)
• As only 2 types of socks exist. Let us assume for simplicity 5pair are red and 5pair are blue.
  total socks 20 and we have to select 3. then n(s)=20C3
  we can select one pair of red sock and 1 blue sock or 1 pair of blue sock and 1 red sock
  two combination possible
  hence (10C2*10C1+10C2*10C1)/20C3
  ANS 900/1140=15/19
• 9 years agoHelpfull: Yes(0) No(3)
• 6*5*4*3*14c1/18c5
• 9 years agoHelpfull: Yes(0) No(0)
• 6C2*5*4*3+6*5C2*4*3+6*5*4C2*3+6*5*4*3C2/18C5
  =5/17
• 9 years agoHelpfull: Yes(0) No(0)
• {(6*5*4*3c2)+(6*5*4c2*3)+(6*5c2*4*3)+(6c2+5*4*3)}/18c5
  =(360+540+720+900)/18*17*14*2
  =5/17
• 9 years agoHelpfull: Yes(0) No(0)
• can any one explain why the following process is not correct??
  
  (6c1*5c1*4c1*3c1)(14c1)/18c5=10/17
  
• 9 years agoHelpfull: Yes(0) No(0)