Probability
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Q.

There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?

 A.

6/7

 B.

1/8

 C.

3/8

 D.

5/9

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Solution:
Option(C) is correct

Total cases of checking in the hotels $= 4^3$ ways.

Cases, when 3 men are checking in different hotels = $4×3×2=24$ ways.

Required probability: 

\(=\dfrac{24}{4^3}\)

\(=\dfrac{3}{8}\)

Edit: Corrected the typo pointed out byBhargavi Bhat.

Edit: For an alternative solution, check comment by Ram Kr Prajapati.


(6) Comment(s)


ADARSH
 ()

4C1*4C1*4C1*4C0=4*4*4=4^3=64



Ram Kr Prajapati
 ()

Total number of ways in which three people can visit hotel out of four hotel,

$={^4C_1} \times {^4C_1} \times {^4C_1}=64$.

Total number of ways in which each person visit different hotel,

$={^4C_1} \times {^3C_1} \times {^2C_1}=24$ (Since Whenever a person visits a hotel then another person is not allowed to visit this hotel).

So Required Probability,

$P=\dfrac{24}{64}$

$P=\dfrac{3}{8}$


Dheeru
 ()

good explanation..this is simple way to solve


Bhargavi Bhat
 ()

instead of 43 ways it should be $4^3$ ways


Deepak
 ()

Thank you Bhargavi for letting me know, corrected the mistake.


Niranjana Murali
 ()

How did we arrive at 4 power 3 in total cases of cheking?