Question 1 :
What is the chance that (i) non-leap year (ii) leap year should have fifty three Sundays?
Solution :
Number of days in non leap year = 365
= 52 week + 1 day
= 52 Sundays + 1 day
Number of Sundays in a year = 52
Days in a week = { Sunday, Monday, Tuesday, Wednesday, Thursday, Saturday }
n(S) = 7
(i) Let "A" be the event of getting the probability of 53 Sundays in a year.
n(A) = 1
P(A) = n(A) / n(S)
= 1/7
(ii) Let "B" be the event of getting the probability of 53 Sundays in a leap year.
Number of days in leap year = 366 days
= 364 days + 2 days
= 52 Sundays + 2 days
Chance of being 53 Sundays = {(Sun, mon),(mon, tue), (tue, wed), (wed, thur) (thur, fri) (fri, sat) (sat, sun)}
B = {(Sun, mon), (sat, sun)}
n(B) = 2
P(B) = n(B) / n(S)
P(B) = 2/7
Question 2 :
Eight coins are tossed once, find the probability of getting (i) exactly two tails (ii) at least two tails (iii) at most two tails
Solution :
When eight coins are tossed, number of events will be = 2^{8 } = 256
n(S) = 256
(i) Let "A" be the event of getting two tails.
n(A) = ^{ 8}c_{2}
n(A) = (8 ⋅ 7)/(2 ⋅ 1)
n(A) = 28
P(A) = n(A) / n(S)
= 28 / 256
= 7/64
(ii) at least two tails
Let "B" be the event of getting at least two tails.
At least two tails means, we may get 2 tails, 3 tails so on up to 8 tails.
Other wise,
= 1 - P(Getting lesser than two tails)
= 1 - [P(0 tail) + P(1 tail)]
= 1 - [(^{8}c_{0}/256) + (^{8}c_{1}/256)]
= 1 - [(1/256) + (8/256)]
= 1 - (9/256)
= (256 - 9) / 256
= 247 / 256
(iii) at most two tails
Let "C" be the event of getting at most two tails.
At most two tails means, we may get 0 tail, 1 tail and 2 tails.
= P(0 tail) + P(1 tail) + P(1 tails)
= (^{8}c_{0}/256) + (^{8}c_{1}/256) + (^{8}c_{2}/256)
= (1/256) + (8/256) + (28/256)
= 37/256
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Oct 06, 24 05:49 AM
Oct 06, 24 05:44 AM
Oct 04, 24 09:58 AM