Latest TCS Aptitude Question SOLUTION: 6 positive integers are taken at random and multiplied together. Then what is the probability that products ends in an odd digit other than 5?

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

Asked In TCS nikhil pande (11 years ago)

Unsolved Read Solution (13)

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Priya Sirohi  (11 years ago)

possible digits at the end of each no is 1,3,7,9 thus probablity for one is (4/10)
we have 6 numbers so probablity will be (4/10)^6

Is this solution Helpfull? Yes (35) | No (7)  

kalyani dangeti  (11 years ago)

product should be odd...if multiplied by 5 we will get 5 or 0 at end..so let us exclude 5...then remaining numbers= even(2,4,6,8),odd=(1,3,7,9)
odd*odd=odd
even*even=even
odd*even=even
=> all numbers should be odd=number of chance=4^6
all possible chances=9^6
probability=(4/9)^6

Is this solution Helpfull? Yes (24) | No (8)  

ramya  (11 years ago)

can anyone explain in detail

Is this solution Helpfull? Yes (8) | No (2)  

Rudra  (11 years ago)

@NIKHIL PANDE WHAT IS THE EXACT ANS ........DO U KNOW???

Is this solution Helpfull? Yes (7) | No (1)  

Sruthi karuturi  (11 years ago)

product should not end with 5 so the digits possible are 1,3,7,9.
Inorder to satisfy this condition the product should be divisible by 2 and 5.
Hence the probability of a no.to be divisible by 2 & 5 is 6/10(0,2,4,5,6,8).
Hence the prob for ending with 1,3,7,9 is 1-(6/10)=4/10=2/5.
For 6 digits the prob is (2/5)^6=64/15625

Is this solution Helpfull? Yes (6) | No (14)  

Aditya  (11 years ago)

Let numbers are 1,2,3,4,5,6....
out of all cases only (1*3)and (3*3)end with odd digit other then 5
so ans is 1/3

Is this solution Helpfull? Yes (1) | No (5)  

Nirmalram  (11 years ago)

possible digits at the end of multiplication is 1,3,7,9
total(1,2,3,4,5,6,7,8,9,0)
probability that products end with odd other than 5 is (4/10)

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

NUJELLA V S L N KARTHIK  (10 years ago)

in any number the last digit can be 0,1,2,3,4,5,6,7,8,9 .We want the last digit in the product is an odd digit other than 5 i.e; it is any one of the digits 1,3,7,9. This means that the product is not divisible by 2 or 5.The probability that a number is divisible by 2 or 5 is 6/10.
[since , in that case the last digit can be one of 0,2,4,5,6,8]
therefore, the probability that the number is not divisible by 2 or 5 = 1- 6/10 = 4/10 = 2/5
in order that the product is not divisible by 2 or 5 none of the constituent numbers should be divisible by 2 or 5 and its probability = (2/5)^6 = 64/390625

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

satnam singh  (10 years ago)

the event will occurs when all the numbers selected are ending in 1, 3, 7 or 9.
if we take numbers between 1 to 10(both inclusive), we will have a positive occurrence if each of the six numbers selected are either 1, 3, 7 or 9....

the probability of any number selected being either of these 4 = 4/10 (4 possible events out of 10 possibilities)
answer is ( 0.4 )^6....

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

manish arora  (11 years ago)

only 0dd*odd gives odd
so
all 6 no. need to be odd
nd also to eliminate no ending with 5 ..
we cant use no. ednding in 5 as any odd*5=end digit 5;
so odd digits that can be taken r 1,3,7,9
nd tatal digits 10
so ans 4/10

Is this solution Helpfull? Yes | No (4)  

rajkumar mukherjee  (10 years ago)

total no. of cases=total no. of odd digits that might be the last digit (1,3,5,7,9)=5
total no. of favorable cases=odd digits excluding 5=(1,3,7,9)=4;
so,probability=4/5;
pls provide optiions;@ nikhil pande

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Kritika Goyal  (8 years ago)

Out of ten positive integers, there are 5 odd and 5 even integers. Odd integers other than 5 are 1,3,7,9.
So, probability of selecting any odd integer other than 5 is given as : 4/10
As the selected integer can repeat itself, the answer is (4/10)^6

Is this solution Helpfull? Yes | No (1)  

mohan  (6 years ago)

The event will occurs when all the numbers selected are ending in 1, 3, 7 or 9.
If we take numbers between 1 to 10 (both inclusive), we will have a positive occurrence if each of the six
numbers selected are either 1, 3, 7 or 9.
The probability of any number selected being either of these 4 is 4/10 (4 positive events out of 10
possibilities)
[Note: If we try to take numbers between 1 to 20, we will have a probability of 8/20 = 4/10. Hence, we
can extrapolate up to infinity and say that the probability of any number selected ending in 1, 3, 7 or 9 so
as to fulfill the requirement is 4/10.]
Hence, answer = (0.4)^6

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