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 probability that a monkey has seen 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%

• B. 39%
  
  p(b)= 0.6
  p(m)= 0.65
  
  as p(b) & p(m) are independent event so
  p(b ∩ m) = p(b)*p(m) = 0.6 * 0.65 = 0.39 = 39 %
• 10 years agoHelpfull: Yes(30) No(14)
• assume that there are 100 monkey from that p(a) is .6 mean
  p(a)= no of monkey who seen banyan tree/total monkey
  .6 = a/100
  a = 60
  60 monkey has seen banyan tree and p(b) is .65 mean 65 monkey has seen mango tree therefor
  monkeys seen both tree are = 60+65-100
  = 25
  that mean 25% of total monkeys
• 10 years agoHelpfull: Yes(23) No(12)
• Rakesh I am not assuming the same. But there is no reason to assume that seeing a banyan is independent from seeing a mango tree. Especially given the wording "minimum fraction of monkeys" which tells you that it will not be a unique answer.
  
  If I say "In a class 80% passed in Physics and 70% passed in Chemistry", will you just assume that 56% must have passed in both? You can only do that if assured that passing in Physics and passing in Chemistry are independent events, right? It might be that the 20 who failed in Physics all also failed Chemistry.
• 10 years agoHelpfull: Yes(8) No(5)
• minimum p(A & B)--->maximum p(A + B)
  suppose total monkey=100;
  max value of p(A + B)=100;
  p(A)=60;
  p(B)=65;
  so p(A & B)=60+65-100=25
  SO 25%
• 9 years agoHelpfull: Yes(5) No(0)
• there are total 1 billion monkeys in the zoo out of which the probability that a monkey in the zoo has seen a banyan tree is 0.6 i.e. 6/10=6000000 for 1billion
  similarly 6500000 monkeys has seen mango tree
  from set theory the min. no of monkeys who have seen both the trees are 2500000 (try to make the vein diagram)hence the min. %age of monkeys who have seen both the trees will be:
  (2500000/10000000)*100=25%
• 9 years agoHelpfull: Yes(3) No(1)
• A.25%
  
  As minimum % is asked so, among 100 monkeys -> 60 saw banyan tree hence 40 monkeys left.
  Now for minimum monkeys who hv seen both the trees ,these 40 should be among 65 Mango tree monkeys .. Finally we are with 25 monkeys who have seen both trees.
• 9 years agoHelpfull: Yes(1) No(2)
• 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%
  
  
• 9 years agoHelpfull: Yes(1) No(1)
• since both events are independent to each other.so ans is .6*.65=39%
• 10 years agoHelpfull: Yes(0) No(1)
• The answer is b
  39%
• 10 years agoHelpfull: Yes(0) No(1)
• 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%
• 6 years agoHelpfull: Yes(0) No(0)