Ravi has a bag full of 10 Nestle and 5 Cadbury chocolates. Out of these, he draws two chocolates. What is the probability that he would get at least one Nestle Chocolate ?

1) 19/21
2) 3/7
3) 2/21
4) 1/3

• Total number sample space two chocolates can be drawn in the way of n(S) = 15C2= 105
  At least one to be nestle.. n(E) = 10C1 X5C1 + 10C2 X 5C0 = 95
  P(E) = 95/105 = 19/21
• 10 years agoHelpfull: Yes(35) No(0)
• (10c1*5c1+10c2)/15c2=19/21
• 10 years agoHelpfull: Yes(9) No(1)
• it can b done as follows....
  (1Nestle and 1Cadbury) or (1Cadbury and 1Nestle) or (2Nestle) i.e.
  (10/15 * 5/14) + (5/15 * 10/14) + (10/15 * 9/14) on solving....
  5/21 + 5/21 + 3/7 which on solving will give ans as 19/21 :)
  
• 10 years agoHelpfull: Yes(6) No(0)
• a)19/21
  is right anwer
• 10 years agoHelpfull: Yes(2) No(3)
• 10C2/15C2=3/7
  
• 10 years agoHelpfull: Yes(1) No(7)
• (10c2)+(10c1*5c1)=95/105=19/21
• 10 years agoHelpfull: Yes(1) No(0)
• A)19/21
  
  
  
  we use combination
• 10 years agoHelpfull: Yes(1) No(0)
• the probability of getting atleast one chocolate is
  (10C1*5c1)+10c2/15c2=19/21
• 8 years agoHelpfull: Yes(1) No(0)
• 
  
  Have any one got the online test result of Cognizant through amcat held at st.peters engg. College, Hyderabad fr 2014 batch on 29th december
  
  
• 10 years agoHelpfull: Yes(0) No(5)
• [(10c1*5c1)/15c2]+[(10c2/15c2)]=19/21
  
• 8 years agoHelpfull: Yes(0) No(0)
• option 1
  10c1 * 5c1/ 15c2 + 10c2/15c2
  on solving this we get 19/21
• 8 years agoHelpfull: Yes(0) No(0)
• Total probability-no Nestle chocolate=1-5c2/15c2=19/21
• 8 years agoHelpfull: Yes(0) No(0)
• Choosing no nestle chocolate -> choosing 2 cadbury / choosing 2 chocolates from total chocolates
  -> 5 C 2 / 15 C 2
  -> 2/21
  Choosing atleast 1 nestle -> 1 - choosing no nestle
  -> 1- 2/21
  -> 19/21
  ( ans ) 19/21
• 5 years agoHelpfull: Yes(0) No(0)