
16 
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. 
38 

B. 
39 

C. 
40 

D. 
41 
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Correct answer is : B
Explanation probability that a monkey has seen banyan tree p(b)= 0.60
probability that a monkey has seen mango tree 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 %
17 
56 toffees have to divided into A, B, C, D, E so that each one gets minimum of 10 toffees. Find the number of ways? 

A. 
10C4 

B. 
12C4 

C. 
11C4 

D. 
10C5 
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Correct answer is : A
Explanation first we will distribute 10 toffees to each of 5 persons i.e total 50 toffees ,
now remaining toffees are 6 which we have to distribute among 5 person so this can be possible in
^{m+r1} C_{ r1} =>^{6+51}C_{51}=^{10}C4
18 
Ram and Shyam have a cube each. Ram paints 4 faces of cube with red colour and rest with blue colour. Ram asks Shyam to paint his cube as well with some sides red and some sides blue. They now start rolling the cubes simultaneously. After doing this for very long Ram observes that probability of both the cubes coming up with same colour is 1/3. How many faces of his cube did shyam paint red. 

A. 
0 

B. 
2 

C. 
3 

D. 
4 
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Correct answer is : A
Explanation for the shyam's cube, let x be the no.of faces painted in red and y be the no.of faces painted in blue.
getting red in 1st cube=4/6=2/3
getting red in 2nd cube= x/(x+y) so probability for red= (2/3)*(x/(x+y))............(1) similarly probability for blue=(1/3)*(y/(x+y))..................(2) given that (1)+(2)=1/3 by solving we get x=0
19 
what is the probability of dividing a randomly selected number (x^4 + 1) by 5,where as x is a four digit number? 

A. 
1 

B. 
0 

C. 
0.9 

D. 
0.2 
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Correct answer is : B
Explanation for (x^{4} + 1) to be divisible by 5, than it should have 0 or 5 as the last digits.
unit place of x^4 will be any of these
0^4=0;1^4=1;2^4=6;3^4=1;4^4=6;5^4=5;6^4=6;7^4=1;8^4=6;9^4=1 hence if we replace X by any of d no: from 0 to 9, non of d no: contain last digit 0 or 5. hence probability is 0.
20 
From a pack of 52 playing cards,three cards are drawn random. Find the probability of drawing a king,a queen and jack. 

A. 
16/5525 

B. 
16/525 

C. 
6/5525 

D. 
16/552 
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Correct answer is : A
Explanation we have 4 of king, queen. and jack so, (^{4}C_{1} * ^{4}C_{1} * ^{4}C_{1})/(^{52}C_{3})=16/5525
