Latest Elitmus Aptitude Question SOLUTION: 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

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(#M40018189) ELITMUS question ELITMUS MATHS PUZZLE Keep an EYE

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%

Asked In Elitmus pavan (11 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

vinod  (11 years ago)

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%

Is this solution Helpfull? Yes (23) | No (14)  

Saiyam Behl  (10 years ago)

.65-.4=.25
.25*100=25%

Is this solution Helpfull? Yes (16) | No (8)  

naga sarath chand  (10 years ago)

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.

Is this solution Helpfull? Yes (4) | No   

Dinesh Gamot  (8 years ago)

P(a)*p(b)
0.65*0.6=0.39
hence answer is 39%

Is this solution Helpfull? Yes (2) | No   

anjan  (8 years ago)

its 25%.
we know P(B)=60%
p(m)=65
p(m&b)=x
using venndiagram,
60-x+x+65-x=100
x=25

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

Tushar Chetan  (5 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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