Probability that number is divisible by 5

please give me the soln. 

3Comments
meet @meet
24 Jan 2016 02:41 am

Solution for without repetition: _ _ _ _  (consider first place start from left side)

here,number should be greater than 5000 so in ----> first place u must put 5 or 7 so u have 2 choices 

second place--->u can put any one of four (0,1,3,(5 or 7 bcz one of them we put in first place) ) so we have 4 choices

third place--->u can put any one of three(bcz three are remaining) so u have 3 choices to put them

forth place---->u can put one of two(remaining two) so u have 2 choices

total no can be formed = 2*4*3*2 =48

now calculate total number which is divisible by 5 and greater than 5000

case 1: first place 5 and last place 0 and remaining two places u have choices to put any of them (1,3,7) so total                      number formed by this case 1*2*3*1=6

case 2:first place 7 and last place 0 or 5(two choice) and remaining place u have 2*3 choice so total number formed             by this case are 1*2*3*2=12

total number=12+6=18 (which is divisible by 5)

probability =18/48=6/16=3/8

u can try ur self problem 1 (digit are repeated) 

Tinu Dahiya @tinudahiya
24 Jan 2016 03:36 pm

here,number should be greater than 5000 so in ----> first place u must put 5 or 7 so u have 2 choices 

forth place---->u can put one of two(remaining two) so u have 2 choices

after filling the place 1st and last we left 3 options, so 

second place--->so we have 3 choices

third place--->(two options are remaining) so u have 2 choices to put them

total no can be formed = 2*3*2*2 =24

see this soln. @meet

Arvind Rawat @arvind.rawat
2 Jun 2016 12:52 pm

your logic for placing digits in forth place is wrong. If you placed 7 in first position then you have 2 choices for forth place(0, 5). But if you placed 5 in first position then u dont have 2 choices for forth place, you can only place 0 there.

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