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Numerical Ability
Probability
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?
Read Solution (Total 13)
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- 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 - 11 years agoHelpfull: Yes(35) No(7)
- 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
- 11 years agoHelpfull: Yes(24) No(8)
- can anyone explain in detail
- 11 years agoHelpfull: Yes(8) No(2)
- @NIKHIL PANDE WHAT IS THE EXACT ANS ........DO U KNOW???
- 11 years agoHelpfull: Yes(7) No(1)
- 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 - 11 years agoHelpfull: Yes(6) No(14)
- 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 - 11 years agoHelpfull: Yes(1) No(5)
- 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) - 11 years agoHelpfull: Yes(1) No(3)
- 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
- 9 years agoHelpfull: Yes(1) No(1)
- 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.... - 9 years agoHelpfull: Yes(1) No(1)
- 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
- 11 years agoHelpfull: Yes(0) No(4)
- 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 - 9 years agoHelpfull: Yes(0) No(1)
- 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 - 8 years agoHelpfull: Yes(0) No(1)
- 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 - 6 years agoHelpfull: Yes(0) No(0)
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