Thus, P(A) = m/n
Let us use this formula and solve some interesting probability problems.
Pro 1:Two dice are thrown simultaneously. Find the probability of getting :
a)an even number as the sum
b)the sum of prime number
c)a total of at least 10
d)a doublet of even number
e) a multiple of 2 on one dice and a multiple of 3 on the other
f)same number on both dice
g)a multiple of 3 as the sum
Solution: When two dice are thrown together the sample space S associated with the random experiment is given by
S= { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), ,
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
Clearly total number of events is 36
a) Let A be the event " getting an even number as the sum" i.e 2,4,6,8,10,12as the sum. Then,
A= { (1,1),(1,3),(3,1), (2,2),(1,5),(5,1),(3,3),(2,4),(4,2),(3,5),(5,3),(4,4),(6,2)(2,6),(5,5),(6,4),(4,6),(6,6)}
Favourable number of elementary events = 18
So, required probability = 18/36 =1/2
b) Let A be the event " getting the sum as a prime number" i.e 2,3,5,7,11 as the sum.Then,
A ={(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(6,5),(5,6)}
Favourable number of elementary events = 15
So, required probability = 15/36 = 5/12
c)Let A be the event of "getting a total of at least 10" i.e 10,11,12.Then,
A={(6,4),(4,6),(5,5),(6,5),(5,6),(6,6)}
Favourable number of elementary events= 6
So, required probability=6/36=1/6
d)Let A be the event of getting a double of an even number.Then ,
A={(2,2),(4,4),(6,6)}
Favourable number of elementary events= 3
So, required probability=3/36=1/12
e) Let A be the event of "getting a multiple of 2 on one dice and a multiple of 3 on the other dice".Then,
A={(2,3),(2,6),((4,3),(4,6),(6,3),(6,6),(3,2),((3,4),(3,6),(6,2),(6,4)}
Favourable number of elementary events= 11
So, required probability=
f) Let A be the event of "getting the same number on both the dice."Then,
A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}
Favourable number of elementary events= 6
So, required probability=6/36 = 1/6
g) Let A be the event " getting a multiple of 3 as the sum" i.e 3,6,9,12 as the sum.Then,
A = {(1,2),(2,1),(1,5),(5,1),(2,4),(4,2),(3,3),(3,6),(6,3),(5,4),(4,5),(6,6)}
Favourable number of elementary events= 12
So, required probability=12/36 = 1/3
Some interesting probability problems:
Pro 2: Find the probability that a leap year will contain 53 sundays.
Sol: In a leap year there are 366 days.
366 days = 52 weeks and 2 days
Thus, a leap year has always 52 Sundays.The remaining 2 days can be:
(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.
If S is the sample space associated with this problem , then S consists of the above seven points.
The total number of elementary events =7
Let A be the event that a leap year has 53 Sundays.In order that a leap year, selected at random, should have 53 Sundays, one of the "over" days must be a Sunday.This is can be in any one of the following two ways
(i) Sunday and Monday or (ii)Saturday and Sunday
Favourable number of elementary events =2
Hence, required probability = 2/7
Some more interesting probability problems:
Pro 3: The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e from 0 to 9.The lock opens with a sequence of four digits with no repeats.What is the probability of a person getting the right sequence to open the suitcase.
Sol: There are 10C4 x 4! = 5040 sequence of 4 distinct digits out of which there is only one sequence in which the lock opens
Therefore, required probability = 1/5040
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