An integer is chosen at random from the first 80,000 positive integers . What is the probability that the integer chosen is divisible by atleast one of the numbers 2,4,6,7
Answers
Answered by
0
Answer:
Let A=the integer is divisible by 6
A=6,12,18,....198
198=6+(n−1)6
⇒n=33
So, P(A)=
200
33
B=the integer is divisible by 8
B=8,16,24,....200
200=8+(n−1)8
⇒n=25
So, P(B)=
200
25
Both are divisible by 6 and 8 both
A∩B=24,48,.....192
192=24+(n−1)24
⇒n=8
So, P(A∩B)=
200
8
∴P(A∪B)=P(A)+P(B)−P(AB)
=
200
33
+
200
25
−
200
8
=
4
1
Answered by
0
Answer:
The number of numbers which are divisible by 6 which are less than 200 is 33.
The number of numbers which are divisible by 8 which are less than 200 is 25.
The number of numbers which are divisible by both 6 and 8 which are less than 200 is 8.
Therefore the number of numbers which are divisible by 6 or 8 is 33+25−8=50.
The probability is 50/200 = 1/4.
Similar questions