seven white balls and three black balls are randomly placed in a row. find the probability that no two black balls r placed together.

• @Syed Zohair Abbas
  All your explation is correct.
  Just we can't fix position of 7 balls ourself..So they should be first arranged in 7! ways then we have 8 positions for the 3 Black balls
  So favorable conditions= (7!*8C3)
  Hence the probability= (7!*8C3)/10!
• 11 years agoHelpfull: Yes(30) No(10)
• 7!*8p3/10!=7/15 8p3 because after selection we also have to arragnge those black balls
  
• 11 years agoHelpfull: Yes(27) No(4)
• when 7 white balls are placed there are 8 vacant places for the red balls to be kept making sure that no two black balls are together. it can be done in 8C3 ways. and since the probability of not having two black balls together is asked then it should be
  
  P(E)= 8C3/10!
  
  why 10!? it is because of in the total of ten balls can be placed in 10x9x8x7x6x5x4x3x2x1 = 10!
  
  Thank You
• 11 years agoHelpfull: Yes(20) No(17)
• 8c3/10c3
  First fix the seven white balls, then you have eight places for the black balls to choose from... and 8C3 is the number of combinations you can have ( how the places are chosen)
  Total number of arrangements of the 10 balls is 10C3
• 11 years agoHelpfull: Yes(9) No(4)
• 7 white balls, so 8 vacant places. 3 black balls can be placed in 8 places in 8C3/3! ways.
  total (7+3)=10 balls can be arranged in 10!/(3!7!) ways.
  so P=(7!*8C3)/10!
• 11 years agoHelpfull: Yes(9) No(1)
• probability=8c3/10c3
• 11 years agoHelpfull: Yes(9) No(2)
• seven white balls that can be arranged in 7! ways . now we have 8 gaps in between to select 3 black balls so 8c3. these 3balls can be arranged again in 3!ways so the required probability (7!*8c3*3!)/10!
• 11 years agoHelpfull: Yes(7) No(1)
• and if we consider same color balls are identical then ans will be 8c3/10c3...10c3 because 10!/7!*3!=total number of arrngmnts
• 11 years agoHelpfull: Yes(4) No(0)
• let us take the event that is opp to required condition ... i.e., all balls are togethr ... so ..., 3 black balls as a packet and ..remaining 7 white as individual .. so ..totally ... 3! 8! chances ...out of 10! ...
  so
  p(E~)=3!8!/10! =1/15 => so /required event is p(e)=1-1/15=14/15
• 11 years agoHelpfull: Yes(3) No(2)
• i think it should be 8C3/(10!/ 7!*3!)
  as the 7 red balls and 3 white balls are identical..so we should devide it
• 11 years agoHelpfull: Yes(3) No(0)
• @Syed Zohair Abbas
  
  It would be (7!*8C3)/ 10!
• 11 years agoHelpfull: Yes(2) No(0)
• @Syed Zohair Abbas
  Ok if you take these balls as identical then the total permutation can also not be 10!.
  It would be 10!/7!*3!
• 11 years agoHelpfull: Yes(2) No(1)
• (8c3*3!*7!)/10!
  
• 11 years agoHelpfull: Yes(2) No(2)
• 7!*8p3/10 is the ans as i think
• 11 years agoHelpfull: Yes(1) No(0)
• 7!*8p3/10!=7/15 8p3 because we need to placed(arrange) them not select
  
  10! becoz total balls are 10
• 10 years agoHelpfull: Yes(1) No(0)
• white balls are place first in 7! ways _w_w_w_w_w_w_w_
  8 places are there for 3 black balls, so it wil be 8c3*3!
  so favourable condition = 7!*8c3*3!
  probability=(7!*8c3*3!)/10!
  
  
• 9 years agoHelpfull: Yes(1) No(0)
• Sorry, I apologize. Its not red rather its black in the starting sentence
• 11 years agoHelpfull: Yes(0) No(0)
• please justify your correction. any correction in case of faulty answer is expected
  
  @tanu
• 11 years agoHelpfull: Yes(0) No(0)
• KoDALI KRANTHI
  your explaination is better
• 11 years agoHelpfull: Yes(0) No(2)
• but is their any need of arranging balls which are identical. there is no need of arranging those seven identical balls. for such a problem if you can please refer to arun sharma's quants question paper. its given there
  
  @tanu
• 11 years agoHelpfull: Yes(0) No(0)
• there are 7 white balls so, no of vacant place where, any of the 3 balls can be placed. and total ways of placing these 7 white and 3 blacks is 10!.
  so, the probability=8c3/10!
  
• 11 years agoHelpfull: Yes(0) No(1)
• 8P3*7! since places r 8 and 3 blck ball can b arrange in 8P3 ways and white ball can b arrnge in 7! ways
• 11 years agoHelpfull: Yes(0) No(0)
• ans 7! *(8P3) as we will first arrange all white balls so 7! and then one one space on each side of white ball 8 spaces are there and we have to arrange onl 3 balls so 8p3
• 11 years agoHelpfull: Yes(0) No(0)
• i think
  7 balls can be arranged in 7! ways.
  8 vacant places for 3 black balls...
  they can be arranged in 8c3 ways
  total=8c3 * 7!
• 11 years agoHelpfull: Yes(0) No(4)
• 8c3 / 10 Ans.
• 11 years agoHelpfull: Yes(0) No(2)
• 1-(2/10!)=.99
• 9 years agoHelpfull: Yes(0) No(0)
• in this question we have 7 white balls,so we will have 8 spaces to fill 3 black balls.
  Hence the ans is 7/15.
  This question is from arun sharma
  
• 9 years agoHelpfull: Yes(0) No(0)
• we have 7 white balls which is placed in 7p7 ways
  now after placing 7 balls we have 8 spaces and 3 black balls which can be arranged in 9p5 ways
  total sample space (7+3)p(7+3) = 10p10
  probability is (7p7*8p3)/(10p10) = 1/772200
• 9 years agoHelpfull: Yes(0) No(0)
• Ans is 7/90
• 9 years agoHelpfull: Yes(0) No(0)
• after placing 7 balls 8 places will be there where you can put the 3 black balls.
  as : _w_w_w_w_w_w_w_
  so 7 white balls can be arranged in 7! ways and total 10 balls are arranged in 10! ways
  so the probability will be:
  [(7!*8C3)/10!]....
• 9 years agoHelpfull: Yes(0) No(0)
• 7 balls can be arranged in 7! ways then we get 8 places for 3 balls
  selection of 3 places for 8 balls is 8C3 after that this 3 balls can be arranged in 3! ways hence
  7! . 8C3. 3! = 7! . 8P3 ways
  thus total propability will be [ (7! . 8P3 )/10!] (as the total no. ways of arranging is 10!
  hence answer is [ (7! . 8P3 )/10!] =7/15 .
• 9 years agoHelpfull: Yes(0) No(0)