A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
a)1/22
b)3/22
c)2/91
d)2/77

• the total sample space=15c3=455
  now for the red color arrangement possible for all red=5c3=10
  again p=n(E)/n(s)=10/455=2/91 thus option C
  
• 9 years agoHelpfull: Yes(12) No(0)
• c)2/91
  
  
  The probability that all of them are red, is = (5/15)*(4/14)*(3/13)= 2/91 ... option C
• 12 years agoHelpfull: Yes(9) No(7)
• A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
  
  A.
  1
  22
  B.
  3
  22
  C.
  2
  91
  D.
  2
  77
  Answer: Option C
  
  Explanation:
  
  Let S be the sample space.
  
  Then, n(S) = number of ways of drawing 3 balls out of 15
  = 15C3
  = (15 x 14 x 13)
  (3 x 2 x 1)
  = 455.
  Let E = event of getting all the 3 red balls.
  
  n(E) = 5C3 = 5C2 = (5 x 4) = 10.
  (2 x 1)
  P(E) = n(E) = 10 = 2 .
  n(S) 455 91
• 5 years agoHelpfull: Yes(0) No(0)