The bacteria has the probability of split into 3 and probability to die is 1/3rd of the total bacteria.Let the probability is P.Some of them survived with probability 1/5.Then which among the following relation is true?
a)P=1/3+1/5*3
b)P=1/5*(1/8-3)

• here prob for dying = 1/3
  hence prob of survival = 1/5
  if they survive, they will be splited into 3 more bacteria
  hence, p = (1/3) + (3* 1/5)
• 13 years agoHelpfull: Yes(24) No(3)
• ans is a) p=1/3+1/5*3
  probablity to die is 1/3 or if it is survived the probablity is 1/5 and if it get survived it is tripled.
• 13 years agoHelpfull: Yes(5) No(0)
• can u plz explain this solution.
• 14 years agoHelpfull: Yes(2) No(2)
• P=1/3+3*(1/5)
• 13 years agoHelpfull: Yes(2) No(0)
• ans is (a)
• 13 years agoHelpfull: Yes(2) No(0)
• b option is correct
• 13 years agoHelpfull: Yes(1) No(9)
• probability of die is 1/3 or its get tripled the probability of tripled is 2/3*p*p*p thats 2/3*p^3 so both the probability combined as P=1/3+(2/3*p^3)
• 13 years agoHelpfull: Yes(1) No(1)
• b)P=1/5*(1/8-3)
• 14 years agoHelpfull: Yes(0) No(31)
• 1/3 is the bacteria's probability.
  after generating the 3 bacteria the bactria's probability is 1/5.
  so total probability is (1/5 +3*(1/5))
• 13 years agoHelpfull: Yes(0) No(5)
• a is the ans
  
• 8 years agoHelpfull: Yes(0) No(0)