Latest Elitmus Aptitude Question SOLUTION: Q. In a chimpoo zoo there are 1 billion monkeys. The probability of monkeys seen in mango trees is 0.6 and probability of monkeys seen in banana tree is 0.65, then what is minimum

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(#M40031091) ELITMUS Numerical Ability question quant Keep an EYE

Q. In a chimpoo zoo there are 1 billion monkeys. The probability of monkeys seen in mango trees is 0.6 and probability of monkeys seen in banana tree is 0.65, then what is minimum percentage of monkeys seen in both the trees??

Options
a) 25%
b) 39%
c) 40%
d) 60%

Asked In Elitmus Vijay Singh Adhikari (10 years ago)

Unsolved Read Solution (11)

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Read Solution (11) :

kishor Mali  (10 years ago)

for mutually independent events:
P(A and B)=P(A).P(B)
P(A or B)=P(A)+P(B)
event A=monkey seen in mango tree
event B=monkey seen in banana tree
as given: both the tree means (A and B) not or
so P(A and B)=(0.6)*(0.65)=0.390=39%

Is this solution Helpfull? Yes (26) | No (5)  

ranjeet singh  (10 years ago)

both tree=p(banana tree)*p(mango tree)
=0.65*0.6=0.390
=39% ans

Is this solution Helpfull? Yes (7) | No (5)  

Liti  (9 years ago)

how to know that both are independent events??

assuming total 100 monkeys,

seen banyan tree= 60,

seen mango tree= 65

so according to set theory, monkeys seen both trees= 60+65-100=25
so ans is 25%

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N Arun Singh  (10 years ago)

.65*.6=.39 i.e 39%
@vijay the events are mutally independent evens

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vishal singh thapa  (10 years ago)

propability("monkeys at mango trees" AND "monkeys at banana trees")= propability(monkeys at mango trees)*propability(monkeys at banana trees)
=0.6*0.65
=0.39
which indicates that 39% of monkeys are in both the trees.

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Vijay Singh Adhikari  (10 years ago)

thanx kishor nd arun..:)

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ranjeet singh  (10 years ago)

because both are independent events thats why can multiply both the probability..

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rahul kumar  (9 years ago)

P(0.6 and 0.65)=0.6*0.65=0.39*100=39%

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liril shrma  (8 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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nitu kumari  (7 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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Vijay Singh Adhikari  (10 years ago)

@ ranjeet ...Can u explain ur solution .

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