1. If integers m and n are chosen at random between 1 and 100, then the probability that a number of the form 7m+7n is divisible by 5 is,
a.1/4 b.1/7 c.1/8 d.1/49
2. A six faced fair dice is thrown until 1 comes, then probability that 1 comes in even no. of trials is,
a.5/11 b.5/6 c.1/6 d.6/11
3.Wats the prob. that a leap year contains 53 sundays?
For 1st n 2nd, pls provide ur method in short.
62think like this what is remainder of 7n on division by 5
n=1 rem = 2
n=2 rem = 4
n=3 rem = 3
n=4 rem = 1
then it repeats...
for the above to hold...
should m be 1 then n is 3
should m be 2 then n is 3
should m be 3 then n is 3
should m be 4 then n is 3
so the probability is 1/4th!!
622. A six faced fair dice is thrown until 1 comes, then probability that 1 comes in even no. of trials is,
a.5/11 b.5/6 c.1/6 d.6/11
P(m) is the prob of getting 1 in the mth throw
P(1) = 1/6
P(2) = (5/6)1/6
P(3) = (5/6)21/6
P(4) = (5/6)31/6
P(5) = (5/6)41/6
P (even) = (5/6)1/6+ (5/6)31/6 + (5/6)51/6 + ........
P (even) = (5/6)1/6{ 1+ (5/6)2 + (5/6)4 + ........ }
this is (5/6)1/6/ { 1 - (5/6)2 }
this is 5/11
623.Wats the prob. that a leap year contains 53 sundays?
Leap year has 366 days = 7 x 52 + 2
Leap year having 53 sunday.. means that the first two days must be a sunday!
so 2/7 !!
13. 2/7
A leap year has 366 days, therefore 52 weeks i.e. 52 Sunday and 2 days.
The remaining 2 days may be any of the following :
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
For having 53 Sundays in a year, one of the remaining 2 days must be a Sunday.
n(S) = 7
n(E) = 2
P(E) = n(E) / n(S) = 2 / 7
1also we can think like this...
71 ends in 7 and 73 ends in 3 whose sum totals to 10
thus the pairs are (1,3) (2,6).........
the other pairs being (2,4) where the sum is also 10 thus taking the pairs and by applying airthametic progression we'll get the answer.
1
i know this is totally unwanted
but i this is the technical solution...
7°=1
7^1=7
72=49
73=__3
74=__1
and this continues...
so all that we need to do is pick up either a number of the form 4K+2
and 4K or of the form 4K+1 and 4K+3
from 1-100 there are 25 numbers each of type 4K,4K+1,+2 and 4K+3
so event space= 25*25*2 +25*25*2
sample space=100*100
n(E)/n(S) =1/4
hence A)
i kno this was unwanted, but just becoz i worked on this im puttin this up...
hope i dont offend anyone coz i didnt read that "Shortest solution thingy"
