[chandler]

there you go :-) that's the correct ans -

41c20(1c1)^20(4c1)^21

[chandler]

sorry for cluttering, I meant -

no of favourable events - 41c20 * (1c1)^20 * (4c1)^21

total no of events - 5^41

I came up with this prob last night and this is what I thot the solution shud be :-)

[chandler]

probability of getting at least 20 correct -

a = summation(I=20 to 41) { (41cI) * (1c1)^I * (4c1)^(40-I) } / 5^41

probability of getting at most 19 correct -

b = summation (I=1 to 19) { (41cI) * (1c1)^I * (4c1)^(40-I) } / 5^41

probability of getting none of them correct -

c = 4^41 / 5^41

Finally -
a+b+c = 1

Please correct me if I am wrong :-)

[rktripathy]

i completely agree with the solution

one more problem
in a country no two people have same sets of teeth.what is the population of the country.(assume a man has 32 no of teeths)
what will be the solution

[made_for_iims]

Originally Posted by rktripathy

i completely agree with the solution

one more problem
in a country no two people have same sets of teeth.what is the population of the country.(assume a man has 32 no of teeths)
what will be the solution

2^32..............

0 to 11111(31 times).....{in binary representation}

so in all 2^32

[rktripathy]

have u included the person having all the 32 teeths or not

[chandler]

or shud it be 0 to 1111(32 times)

i.e 2^33

[chandler]

oops 0 to 11111...(32 times) = 2^32

[Eccentric]

Originally Posted by rktripathy

i completely agree with the solution

one more problem
in a country no two people have same sets of teeth.what is the population of the country.(assume a man has 32 no of teeths)
what will be the solution

What does same set means here.
I think the tooth can be subtlely different. Even if all people had only one tooth, I can have infinite population each person having different tooth!
Set usually means specified size/shape of 2 molars, 4 premolars, inscisor, grinder etc so that it fits in your jaw bones.
SO different sets means size and shape are different with same combination.

[suprovo2000]

q1. The probability that a graduate student being male is 0.25 and that being female is 0.75. The probability that a male student passes the course is 0.7 and that a female student does it is 0.80. A student selected at random is found to have completed the course. What is the probability that the student is (i) male and (ii) female?



q2. Three members X, Y and Z of a private club have been nominated for the office of the president. The probability the Mr. X will be elected is 0.3, the probability that Mr. Y will be elected is 0.5, and the probability that Mr. Z will be elected is 0.2.Should Mr. X be elected, the probability for an increase in membership fee is 0.8? Should Y or Z be elected, the corresponding probability for an increase in fee is 0.2 and 0.3?If fee has been increased, what is the probability that (i) Mr. X was elected president of the club? (ii) Mr. Y was elected president of the club?