probablity

Q1 if n postiive integers are taken at random and multiplied together and pn is probablity that last digit of product is 2,4,6,8.Find pn and p4

Q2 2 integers r,s are drawn one at a time without relacememnt from set 1,2,3...n.If pk=P(r≤k l s≤k).Find 4p7 if n=25

9 Answers

1
rickde ·

1)
all we requite is that atleast one of the n integers must be even n none of them must be divisivle by five
because only 2x5 will end in 0 rest all will give 2,4,6 or 8

prob integer is even =1/2
prob integer is not divisible by five and two=1-1/5-1/2+1/10=2/5

so probability = \sum_{r=1}^{n}{{n\choose r}\left(\frac{1}{2} \right)^r\left(\frac{2}{5}\right)^{n-r}}

=\left(\frac{1}{2}+\frac{2}{5} \right)^n - {n\choose 0}\left(\frac{2}{5}\right)^n

=\left(\frac{7}{10} \right)^n - {n\choose 0}\left(\frac{2}{5}\right)^n

This is wrong as cases become too complicated n selection not required

1
rickde ·

i have taken that the set of numbers from which we r choosing is not limited...i.e. all natural numbers

1
rickde ·

cud u elaborate q2 ...the symbols r not clear

24
eureka123 ·

ans1= (4/5)n-(2/5)n

and whats not clear in 2 ??

1
rickde ·

wat do u mean after the IF statement......n wat 4p7

1
rickde ·

got the first one......

we first include all possible numbers....
we see that in no way numbers divisible by five...(as it will then end in 0 or 5)
all other numbers are permitted

we see that 4 out of every 5 numbers is not divisible by 5

so \left(\frac{4}{5}\right)^n

but in this we have included a case where all numbers end with either 1 or 3.....but atleast one even is required

so we have to subtract
\left(\frac{2}{5}\right)^n cases

as 2 out of every 5 number ends by 1 or 3

hence the ans
\left(\frac{4}{5}\right)^n-\left(\frac{2}{5}\right)^n

24
eureka123 ·

looks ok..thx

1
rickde ·

arre eure ...clarify q2 na....

24
eureka123 ·

whats not clear ??
its 4. p7 where pk=P(r≤k l s≤k)

** l denotes ""such that ""

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