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(#M40022758) TCS Numerical Ability question Probablity Keep an EYE

6 positive number are taken at random and multiplied together. Then what is the probability that the product ends in an odd digit other than 5

Asked In TCS aamir alam (10 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

Rahul Agrawal  (10 years ago)

Total no of ways is 10
Reqired selections
(i.e the selected digit must end with 1,3,7 or 9)
= 4
hence probability= 4/10= 2/5

Is this solution Helpfull? Yes (12) | No (4)  

shouvik ghosh  (10 years ago)

number ending with 1,3,7,9 should be taken
so (2/5)^4=16/625

Is this solution Helpfull? Yes (5) | No (15)  

Nipun  (10 years ago)

Answer is (4/10)^6 ....
Total Ways of getting any positive number = 10^6
To get an odd number (not having 5 in unit's place) we'll need - 1,3,7,9 and each of which will appear in 4^6 ways in unit's place.

so answer - (4/10)^6

Is this solution Helpfull? Yes (5) | No   

hemalatha  (10 years ago)

can any one tell me solution

Is this solution Helpfull? Yes (1) | No (11)  

pranjul chandel  (9 years ago)

Ans is ---> 64/15625

Is this solution Helpfull? Yes (1) | No   

aneeshabsoman  (9 years ago)

shouvik ghosh your answer is correct..i checked it with another website
16/625 but i didnt understand how you did it. could you please explain?

Is this solution Helpfull? Yes | No (1)  

Ratan Dutta  (9 years ago)

the unit digit must be odd according to the qs.....
so we hv 1,3,7,9...this 4 odd digit out of 10 positive digits excluding 5....
so probability of avalibilty of odd digits=4/10=2/5
but this is not the ans...bcz we cannot put the 4 digits at a time in unit digits place...its impossible...
so there any one out of the 4 digits can be...
so the probabilty is the no of ways in which that is possible and that is=(2/5)^4=16/625

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