Latest TCS Aptitude Question SOLUTION: 4 men throw a die each simultaneously. Find the probability that at least 2 people get the same number a. 5/18 b. 13/18 c. 1/36 d. 1/2

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(#M40154755) TCS Numerical Ability question Probablity Keep an EYE

4 men throw a die each simultaneously. Find the probability that at least 2 people get the same number
a. 5/18
b. 13/18
c. 1/36
d. 1/2

Asked In TCS shoaib ibrahim (8 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

nilu kumari  (8 years ago)

Explanation:
The number of ways of rolling a dice where no two numbers probability that no one rolls the same number = 6 x 5 x 4 x 3
Now total possibilities of rolling a dice = 64
The probability that a no one gets the same number = 6×5×4×3 / 6^4 = 5/18
So the probability that at least two people gets same number = 1−5/18=13/18

Is this solution Helpfull? Yes (21) | No (5)  

sanjeev kumar  (8 years ago)

b)13/18
P=1-(6*5*4*3)/(6*6*6*6)

Is this solution Helpfull? Yes (5) | No   

tanmay das  (8 years ago)

ans is option B=13/18

Is this solution Helpfull? Yes (2) | No (3)  

puppy  (8 years ago)

6. If 3

Is this solution Helpfull? Yes | No (2)  

  (4 years ago)

Let's first find the probability that all the people get different numbers.

Outcome of each die can be any of the 6 numbers. Therefore,
Total number of outcomes
=
6
4


First die can give any of the 6 numbers
Second die can give any of the remaining 5 numbers
Third die can give any of the remaining 4 numbers
Fourth die can give any of the remaining 3 numbers
Therefore, total favorable outcomes 6×5×4×3

P(all people get different numbers)
=
6
×
5
×
4
×
3
6
4
=
5
18


P(at least 2 people get the same number)
=
1
−
5
18
=
13
18

Is this solution Helpfull? Yes | No   

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