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A natural number is selected from the set of all natural numbers between 61 and 1020 (both numbers inclusive). What is the probability that the number is a multiple of 3 or 4 or 5?
  • a)
    11/20
  • b)
    12/20
  • c)
    14/20
  • d)
    15/20
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A natural number is selected from the set of all natural numbers betw...
The LCM of 3, 4 and 5 is 60.
Thus checking the remainder of 61 w.r.t. 3/4/5 is the same as checking the remainder of 1 w.r.t. 3/4/5.
Similarly 62 is akin to 2, 63 to 3 and so on.
So let us simply look at the 1st 60 numbers.
Divisible by 3 = |60/3| = 20
Divisible by 4 = |60/4| = 15
Divisible by 5 = |60/5| = 12
Divisible by 3 & 4 = |60/12| = 5
Divisible by 3 & 5 = |60/15| = 4
Divisible by 4 & 5 = |60/20| = 3
Divisible by 3 & 4 & 5 = |60/60| = 1
Thus total number of number divisible by 3 or 4 or 5 = (20 + 15 + 12) - (5 + 4 + 3) + 1
= 47 - 12 + 1
= 36
Thus for every 60 numbers, 36 of them will satisfy, hence probability = 36/60 = 12/20.
Now 61-1020 involves 960 numbers which is 16 sets of 60 numbers, hence Probability = 12/20.
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A natural number is selected from the set of all natural numbers between 61 and 1020 (both numbers inclusive). What is the probability that the number is a multiple of 3 or 4 or 5?a)11/20b)12/20c)14/20d)15/20Correct answer is option 'B'. Can you explain this answer?
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