probability

each packet of certain items contain a coupon which is equally likely to bear the letters a,n,s,h,u if m packets r purchased then how to find the probability that the coupons cannot spell ANSHU, how? hint

21 Answers

13
Двҥїяuρ now in medical c ·

@philip

what is "x"?

13
Двҥїяuρ now in medical c ·

Nishant bhaiya.....u dere to explain this to me.....

62
Lokesh Verma ·

sorry dude.. i used recurrsion.. i meant inclusion exclusion..

I have a class now.. will catch you after that :)

That too was very clumsy.. I am not convinced with my own effort on this one... Will give it mroe thought when I have more time :)

13
Двҥїяuρ now in medical c ·

recurrsion relation?[7] plz sir.......

13
Двҥїяuρ now in medical c ·

what is recurrsion relation??sir....plz explain a bit...i didn't get it[7]

for m=5 ...probability=5!/55 is n't it???

62
Lokesh Verma ·

I think it should be

5r - 5.4r + 10. 3r - 10 . 2r + 5 - 1

whole divided by 5r

by recurrsion relation... DId it without giving too much tought.. so may be wrong :)

13
Двҥїяuρ now in medical c ·

Lost thread!!!!

Plz anyone solve it!!!!!!

13
Двҥїяuρ now in medical c ·

P of not getting an A, Probability of Not getting a B....

aren't these independent event???[7]

i myself also had the fallacy with m<5[2]

plz do it for us ...bhiyaa

62
Lokesh Verma ·

abhirup are these independent events?

P of not getting an A, Probability of Not getting a B and so on?

I think there is some mistake

Namely if m<5 answer should be zero or undefined....

13
Двҥїяuρ now in medical c ·

plz see this.....am i right????

13
Двҥїяuρ now in medical c ·

probability of getting no A for m times is (4/5)m
so,probability of getting atleast one A is 1-(4/5)m

so probability of getting atleast one A,N,S,H,U is[1-(4/5)m]5

probability that the coupons cannot spell ANSHU is 1-[1-(4/5)m]5

correct me if i am wrong.....

13
MAK ·

HINT :

Such type of problems can be easily solved in this manner...

P(coupons cannot spell ANSHU) = 1 - P(coupons can spell ANSHU)

first try to find the probability dat d coupons can form the word (or spell the word) ANSHU...

hope this hint is sufficient...!!! [1]

1
pankaj sachan ·

i think answer should be 5(4m/5m)

1
Philip Calvert ·

see abhirup
the easiest method i can think of is
the total no. of combinations of letters possible are 5m
find out the no of combinations that CAN spell anshu let it be x
then 1- x/5m
shud give the answer

13
Двҥїяuρ now in medical c ·

i got

1-[1-(4/5)m]5

am i right???

3
msp ·

ok maq

13
MAK ·

not exactly sankara...!!!

well, i'm not posting d solution juzz because kishan asked only d hint for solving it... he didn't even visit dis thread after posting the question... if he couldn't solve it further, den i'll post my solution... [1]

3
msp ·

maq am i rite.please correct me if i was wrong..

3
msp ·

maq i am getting 1-mC5/5m

13
MAK ·

no abhirup... mC5(1/5)5 can also be ≥ 1 sometimes...

13
Двҥїяuρ now in medical c ·

is it 1- mC5(1/5)m???

not sure

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