Latest Elitmus Aptitude Question SOLUTION: in a zoo, there are 1 billion monkeys. Probability that a monkey has seen a banyan tree is 0.6. Prob. that monkey has seen a mango tree is 0.65. what is the minimum percentage of m

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(#M40152324) ELITMUS Numerical Ability question Probability Keep an EYE

in a zoo, there are 1 billion monkeys. Probability that a monkey has seen a banyan tree is 0.6. Prob. that 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?
OPTIONS- 25%,39%,40%,60%

Asked In Elitmus Souradeep Guhathakurta (8 years ago)

Solved Abdul Bashar Read Solution (14)

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Read Solution (14) :

Abdul Bashar  (8 years ago) Best Solution

prob of seeing banyan tree, P(A)=0.6
prob of seeing mango tree, P(B)=0.65
PROB OF seeing min both the tree, P(A intersaction
using this formula,
p(A union B)= p(A)+p(b)-p(A intersaction B)-------(1)
here we use max prob is 1 , so p(A Union B) =1
putting all values in eqn (1) we get
p(A intersaction B )= 0.25.... i.e. 25%
SO OPTION (B) IS CORRECT

Is this solution Helpfull? Yes (38) | No (4)  

Avinash Sharma  (8 years ago)

Probability that the monkey has seen banyan tree = 0.6
Probability that the monkey has seen mango tree = 0.65
Therefore probability that the monkey has seen the banyan tree and the mango tree = 0.6*0.65 =0.390
Therefore percentage of monkey in the zoo who have seen bothe the trees is = 0.390*100 = 39%
Hence ans b

Is this solution Helpfull? Yes (15) | No (14)  

krishna  (8 years ago)

0.6+0.65=1.25 hence 25%.

Is this solution Helpfull? Yes (4) | No (2)  

jintu  (8 years ago)

I don't think, Abdul Bashar is right. Because there may be monkeys ,who did not see neither banyan nor mango tree. So p(A union B) not equal to 1.
Therefore Avinash's answer is right.

Is this solution Helpfull? Yes (3) | No   

yagyesh  (8 years ago)

p(total) =1 = P(seen banyan tree) + p( seen mango tree) - p(seen both banyan and mango) +p(not seen banyan and not seen mango)
Answer will be p(seen both banyan and mango)
or
p(B intersection M) = p(B) +P(N) + P(!B intersection !N) -1
for p(B intersection M) to be minimum p(!B intersection !N) should be zero because P(B) and P(N) are already given, so for minimum the case is every monkey has seen something..
p(B intersection M) = 0.6+0.65+0-1=0.25
which is 25%

Is this solution Helpfull? Yes (3) | No (1)  

ajit kumar sharma  (8 years ago)

plzzz don't make confusion here....somebody just provide the correct answer......with proper argument....thank you

Is this solution Helpfull? Yes (2) | No (1)  

Anjali Sadhukhan  (8 years ago)

two independent event to be assumed..as nothing such is specified.so option (B) 39% .

Is this solution Helpfull? Yes (2) | No (8)  

pratyaksh  (8 years ago)

Avinash Sharma is correct.
Ans is 39%

See explanations: http://stattrek.com/probability/probability-rules.aspx

Is this solution Helpfull? Yes (2) | No   

parash  (8 years ago)

ans-39% is correct

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pradeep singh rawat  (8 years ago)

ans is 39%.

Is this solution Helpfull? Yes (1) | No (3)  

Kunal Wadhwa  (8 years ago)

Suppose there are 100 monkeys (doesn't matter whether we consider it as 100 or 1 billion because the ans is in percentage)
P (that a monkey has seen banyan tree) = 0.6
this means 60 monkeys have seen banyan tree out of 100 & 40 have not seen banyan tree

P (that a monkey has seen mango tree) = 0.65
this means 65 monkeys have seen mango tree out of 100

For minimum case 40 monkeys that have not seen banyan tree will see mango tree
So minimum 65-40 = 25 monkeys have seen both the trees

Ans : 25 %

Is this solution Helpfull? Yes (1) | No (3)  

Pooja Ghosh  (6 years ago)

So, there are 0.6 billion monkeys who saw a banyan tree and 0.65 billion monkeys who saw a mango tree.
Now the total number of monkeys is the sum of the number of monkeys who saw only a banyan tree, only a mango tree and monkeys who saw both.
Now here we have two categories: monkeys who saw a banyan tree and monkeys who saw a mango tree. Now those who saw both can be included in both the categories. If we add 0.6+0.65=1.25 billion. But here, the monkeys who saw both have been counted twice. If they were counted only once our answer would have been exactly 1 billion. Naturally, 0.25 billion monkeys have been counted twice, i.e. 0.25 billion have seen both. Now, the fraction 0.25 corresponds to 25%. So 25% monkeys have seen both the trees.

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hanendra  (8 years ago)

sorry for mistake in my solution as the correct answer will be 16

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James Anderson  (7 years ago)

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%

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