4 red,6blue,26 green ball. person choose 1 at a time either blue or red. what is probability that he choose red ball

• probability that none of ball is blue or red is=26/36=13/18
  probability of blue or red ball=1/13/18=5/18
  probability of only red ball in blue or red ball=(4/36)/(5/18)=0.4
• 10 years agoHelpfull: Yes(33) No(2)
• Probability that it is neither red nor blue =26/36=13/18
  Probability that it is blue or red=(1-13/18)=5/18
  Probability of red ball=(4/36)/(5/18)=0.4
• 9 years agoHelpfull: Yes(15) No(0)
• prob of choosing 1 red ball is : 4/36=1/9
  prob of choosing blue or red is : 10/36=5/18
  prob of choosing red among blue or red ball choices is : (1/9)/(5/18)=2/5=0.4
• 9 years agoHelpfull: Yes(5) No(0)
• i also got this question in today's TCS test. I answered Correct,
  i.e Ans=0.4
• 10 years agoHelpfull: Yes(3) No(3)
• 0.45
  
  This is the required answer.
• 10 years agoHelpfull: Yes(1) No(8)
• Total No.of red and blue balls =4+6=10
  total number of Events n(S)=10;
  Favorable Events: red balls =4
  then P(E)= 4/10 = 2/5 = 0.4
• 9 years agoHelpfull: Yes(1) No(0)
• 1/54 is the ans
• 10 years agoHelpfull: Yes(0) No(8)
• how .45 preeti can you explain.
  
• 10 years agoHelpfull: Yes(0) No(1)
• 2/3.24/36=2/3
  
• 10 years agoHelpfull: Yes(0) No(0)
• Probability that it is neither red nor blue
  =26/36=13/18
  Probability that it is blue or red=(1-13/18)=5/18
  Probability of red ball=(4/36)/(5/18)=0.45
• 9 years agoHelpfull: Yes(0) No(0)
• we can simply solve it through the subject Analog and Digital's topic MUX nd D-Mux..
  As we know that for minimum 3 output we need at least 2 input..
  and so on as in term of 2's power..
  So for 5 to 8 output we need 3 input(2^3=8)[8 will be maximum output]
  for 9 to 16 output we need 4 input(2^4=14)
  for 17 to 32 output we need 5 input(2^5=32)
  so,Ans wiil be 5 which is not given in the option.
• 9 years agoHelpfull: Yes(0) No(0)
• the person selects either blue or red. so total balls to be considered is only 10. fr0m this only 4 red balls are there. so probability of selecting 1 red ball=4c1/10c1=4/10=.4
• 9 years agoHelpfull: Yes(0) No(0)
• probability that none of ball is blue or red is=26/36=13/18
  probability of blue or red ball=1/13/18=5/18
  probability of only red ball in blue or red ball=(4/36)/(5/18)=0.4
  
• 9 years agoHelpfull: Yes(0) No(1)