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Numerical Ability
Probability
6 positive integers are taken at random & multiplied together. what is the probability that the product ends in an odd digit other than 5?
Read Solution (Total 8)
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- it means at unit position of product only 1,3,7,9 can occur and it is possible only when nos. having 1or 3or 7or 9 at unit position are multiplied.
so total possible cases=10*10*10*10*10*10=1000000
as any of 0,1,2,3,4,5,6,7,8,9 can occur at unit place.
but total desirable cases are 4*4*4*4*4*4=4096 as at unit position of randomly selected nos. only1,3,7,9 are desired so ans is-4096/1000000 or 64/15625. - 11 years agoHelpfull: Yes(15) No(16)
- for the product to b odd,all the numbers have to b odd,thus there are 4 numbers,1,3,7,9.thus the possible cases are 4*4*4*4*4*4
total outcomes are-9*9*9*9*9*9
therefore the probability is-4^6/9^6 - 11 years agoHelpfull: Yes(7) No(2)
- bro..the logic behind 10^6 (not 10^4) is that...u r asked to select 6 random +ve nos. and those 6 selected nos. can hav any of 0,1,2,3,4,5,6,7,8,9 at their unit place..so 10^6 is total no. of cases..hope u got it nw..:)
- 11 years agoHelpfull: Yes(3) No(2)
- @vishnu ... it wont be 10^4...because here they didnt ask for 6 whoe no.s...the asked for 6 positive no.s...and 0 not positive neither negative..so it will be 9^4..else part fine
- 11 years agoHelpfull: Yes(2) No(1)
- any number selected having probability = 4/10=0.4
so the required probability=(0.4)^6 - 9 years agoHelpfull: Yes(2) No(0)
- u r taking it wrong brother..its written there that 6 positive integers..not 6 whole numbers....0 in not a positive no. neither negative...its a nutral no....so u can consider 0 here in this case.....here it will be 9^6....if they asked 6 whole numbers then u would use 10^6..but not in this case...as they said 6 positive integer so it will be 9^6
- 11 years agoHelpfull: Yes(1) No(2)
- the answer is 16/625
- 11 years agoHelpfull: Yes(0) No(0)
- 9*8*7*6*5*3
- 8 years agoHelpfull: Yes(0) No(0)
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