Answer:
0.3
Step-by-step explanation:
Number of ways of selecting r elements from a set of n different elements is given by
C(n,r)=(n r)=n!/(r!(n−r)!)
No of ways of selecting one number out of ten numbers from one to 10 is 10
It can also be calculated using
C(n,r)=(n r)=n!/(r!(n−r)!)
where n = 10 and r = 1
C(10,1)=(10 1)=10!(1!(10−1)!) = 10*9!/ 1!*9! = 10
multiples of 2 in range 1 to 10 are 2, 4, 6, 8 , 10
multiples of 3 in range 1 to 10 are 3, 6, 9
Therefore number which are multiple of 2 and 3 are
2, 3,4,6,8,9,10 ( 7 numbers)
therefore no of ways of selecting multiple of 2 and 3 is 7 ways
number which are not multiple of 2 and 3 in range 1 to 10
1,5,7 (3 numbers)
no of ways of not selecting multiple of 2 and 3 is 3 ways
It can also be calculated using
C(3,1)=(3 1)=3!/(1!(3−1)!)
where n = 3 and r = 1
C(3,1)=(3 1)=3!/(1!(3−1)!) = 3!/(1!(2)!)= 3*2!/ 1!*2! = 2
Therefore no of ways of not selecting multiple of 2 and 3 is 3 ways
probability of not selecting a multiple of 2 or a multiple of 3 = no of ways of not selecting multiple of 2 and 3 /No of ways of selecting one number out of one to 10 = 3/10 = 0.3
