Example 1 :
A year is selected at random. What is the probability that (i) it contains 53 Sundays (ii) it is a leap year which contains 53 Sundays
Solution :
In a ordinary year, we have 365 days
= 52 weeks + 1 day
= 52 Sundays + 1 day
That one day could be = {Sunday, Monday, Tuesday, Wednesday, Thursday, Saturday}
Probability of getting 53 Sundays in ordinary year = 1/7
In a leap year, we have 366 days
= 52 weeks + 2 days
= 52 Sundays + 2 days
That one day could be = {(Sun, mon) (mon, tue) (tue, wed)(wed, thu) (thu, fri) (fri, sat) (sat, sun)}
Probability of getting 53 Sundays in leap year = 2/7
(i) it contains 53 Sundays
Hint in the given question : A year is selected at random
From this, we come to know that we may choose either ordinary year or leap year.
P(getting 53 Sundays)
= P(selecting ordinary year) ⋅ P(53 Sundays in ordinary year)
= (3/4) ⋅ (1/7)
[(A year occurring once in 4 years is know as leap year)
So, probability of being ordinary year = 3/4]
= 3/28 -------(1)
= P(selecting leap year) ⋅ P(53 Sundays in leap year)
= (1/4) ⋅ (2/7)
= 2/28 -------(2)
(1) + (2)
= (3/28) + (2/28)
= 5/28
(ii) it is a leap year which contains 53 Sundays
= P(selecting leap year) ⋅ P(53 Sundays in leap year)
= (1/4) ⋅ (2/7)
= 1/14
Example 2 :
Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?
Solution :
P(Hitting a target by X) = 3/4, P(X') = 1/4
P(Hitting a target by Y) = 4/5, P(Y') = 1/5
P(Hitting a target by Z) = 2/3, P(Z') = 1/3
P(target is damaged by exactly 2 hits)
= P(X'nYnZ) + P(XnY'nZ) + P(XnYnZ')
= P(X')P(Y)P(Z) + P(X)P(Y')P(Z) + P(X)P(Y)P(Z')
= (1/4)(4/5)(2/3) + (3/4)(1/5)(2/3) + (3/4)(4/5)(1/3)
= 8/60 + 6/60 + 12/60
= 26/60
= 13/30
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
May 22, 24 06:32 AM
May 17, 24 08:12 AM
May 14, 24 08:53 AM