In a Chimp Zoo there are 1 billion monkeys. The probability that a monkey in the zoo has seen a Banyan tree is 0.6. The probabilty that a monkey has seen a mango tree is 0.65. What is the minimum percentage of monkeys in the zoo who have seen both the trees?
a 25%
b 39%
c 40%
d 60%

• we know that p(a&b)=p(a)*p(b) for independent events
  here p(a)=0.6
  p(b)=0.65
  so p(a&b)= 0.6*0.65
  = 0.39
  ans 39%
• 11 years agoHelpfull: Yes(23) No(14)
• .65-.4=.25
  .25*100=25%
• 11 years agoHelpfull: Yes(16) No(8)
• o.65*0.60/(100*100) = 39%.is the correct answer.(as both cases should happen).
  p(A intersection B) = p(A)*p(B). as both are independent events.
• 11 years agoHelpfull: Yes(4) No(0)
• P(a)*p(b)
  0.65*0.6=0.39
  hence answer is 39%
• 9 years agoHelpfull: Yes(2) No(0)
• its 25%.
  we know P(B)=60%
  p(m)=65
  p(m&b)=x
  using venndiagram,
  60-x+x+65-x=100
  x=25
• 9 years agoHelpfull: Yes(1) No(3)
• For two independent events P(A and B) = P(A)*P(B)
  
  => P(monkey has seen a banyan tree and monkey has seen a mango tree) = P(monkey has seen a banyan tree) * P(monkey has seen a mango tree)
  
  => P(monkey has seen a banyan tree and monkey has seen a mango tree) = 0.6 * 0.65 = 0.39
  
  So the percentage of monkeys who has seen both banyan tree and mango tree = 39%
• 6 years agoHelpfull: Yes(0) No(0)