. A number is selected at random from the first 1000 natural numbers. What is the probability that it would be a multiple of 5 or 9 1 point
Answers
Explanation:
Given :-
A number is selected at random from the first 1000 natural numbers.
To find :-
What is the probability that it would be a multiple of 5 or 9 ?
Solution :-
Given that :
Total number of natural numbers = 1000
Total number of all possible outcomes = 1000
List of multiples of 5 in first 1000 natural numbers
= 5 , 10 , 15 , ... , 995,1000
First term = (a) = 5
Common difference = (d) = 10-5 = 5
Since they are in the AP ,
The nth term of an AP = an = a+(n-1)d
Let an = 1000
=> 5+(n-1)(5) = 1000
=> 5 + 5n-5 = 1000
=> 5n = 1000
=> n = 1000/5
=> n = 200
Total number of multiples of 5 = 200
Number of favourable outcomes to 5 = 200
We know that
Probability of an event P (E)=
Number of favourable outcomes/Total number of all possible outcomes
Probability of getting a number which is a multiple of 5
= 200/1000
= 1/5
= 0.2
List of multiples of 9 in first 1000 natural numbers
= 9, 18 , 27 , ... , 999
First term = (a) = 9
Common difference = (d) = 18-9=9
Since they are in the AP ,
The nth term of an AP = an = a+(n-1)d
Let an = 999
=> 9+(n-1)(9) = 999
=> 9+ 9n-9 = 999
=> 9n =999
=> n = 999/9
=> n = 111
Total number of multiples of 9 =111
Number of favourable outcomes to 9= 111
We know that
Probability of an event P (E)=
Number of favourable outcomes/Total number of all possible outcomes
Probability of getting a number which is a multiple of 9
= 111/1000
= 0.111
List of multiples of 9×5=45 in first 1000 natural numbers
= 45, 90 , ...,990
First term = (a) =45
Common difference = (d) = 90-45=45
Since they are in the AP ,
The nth term of an AP = an = a+(n-1)d
Let an =990
=> 45+(n-1)(45) = 990
=> 45+ 45n-45 = 990
=> 45n =990
=> n = 990/45
=> n = 22
Total number of multiples of 45 = 22
Number of favourable outcomes to 45 = 22
We know that
Probability of an event P (E)=
Number of favourable outcomes/Total number of all possible outcomes
Probability of getting a number which is a multiple of 45
= 22/1000
= 0.022
Now,.
The probably of a number which is either multiple of 5 or 9
=> (Probability of multiples of 5 + Probability of multiples of 9 )- Probability of multiples of 5 and 9
= 0.2+0.111-0.022
= 0.311-0.022
= 0.289
= 289/1000
Alternative Method:-
Probability of getting a multiple of 5 in first 1000 natural numbers
= (1000/5)/1000
=200/1000
= 0.2
Probability of getting a multiple of 9 in first 1000 natural numbers
= (1000/9)/1000
=~111/1000
= 0.111
Probability of getting a multiples of both 5 and 9 in first 1000 natural numbers
= (1000/45)/1000
=~22/1000
= 0.022
P(A) = 0.2
P(B)=0.111
P(A n B ) = 0.022
We know that
P(AUB) = P(A)+P(B)-P(AnB)
=>P(AUB)=0.2+0.111-0.022
=> P(AUB) = 0.311-0.022
P(AUB) = 0.289
The probably of a number which is either multiple of 5 or 9 is 289/100 or 0.289
Answer:-
The probably of a number which is either multiple of 5 or 9 is 289/100 or 0.289
Used formulae:-
- The nth term of an AP = an = a+(n-1)d
- Probability of an event P (E)=
- Number of favourable outcomes/Total number of all possible outcomes
- P(AUB) = P(A)+P(B)-P(AnB)
Answer:
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