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Probability and Statistics

GATE CSE 2014 Set 2 · Q48

NAT2 marksMedium

Time

The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is _________.

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Solution

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Official answer: 0.26 (±0.001) (±0.001)

Probability and Statistics

The total number of positive integers from 11 to 100100 is N=100N = 100.

Let A2,A3,A5A_2, A_3, A_5 be the sets of integers divisible by 2,3,52, 3, 5 respectively.

Using the Principle of Inclusion-Exclusion:

  • ∣A2∣=⌊100/2⌋=50|A_2| = \lfloor 100/2 \rfloor = 50

  • ∣A3∣=⌊100/3⌋=33|A_3| = \lfloor 100/3 \rfloor = 33

  • ∣A5∣=⌊100/5⌋=20|A_5| = \lfloor 100/5 \rfloor = 20

Pairwise intersections:

  • ∣A2∩A3∣=⌊100/6⌋=16|A_2 \cap A_3| = \lfloor 100/6 \rfloor = 16

  • ∣A2∩A5∣=⌊100/10⌋=10|A_2 \cap A_5| = \lfloor 100/10 \rfloor = 10

  • ∣A3∩A5∣=⌊100/15⌋=6|A_3 \cap A_5| = \lfloor 100/15 \rfloor = 6

Triple intersection:

  • ∣A2∩A3∩A5∣=⌊100/30⌋=3|A_2 \cap A_3 \cap A_5| = \lfloor 100/30 \rfloor = 3

The total number of integers divisible by at least one of 2,3,2, 3, or 55 is:

∣A2∪A3∪A5∣=(50+33+20)−(16+10+6)+3=103−32+3=74|A_2 \cup A_3 \cup A_5| = (50 + 33 + 20) - (16 + 10 + 6) + 3 = 103 - 32 + 3 = 74

Therefore, the number of integers not divisible by 2,3,2, 3, or 55 is:

100−74=26100 - 74 = 26

The required probability is:

P=26100=0.26P = \frac{26}{100} = 0.26

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This Probability and Statistics previous year question is Q48 of the GATE CSE 2014 Set 2 paper. Find more GATE CSE PYQs on Probability and Statistics in Engineering Mathematics, or browse all GATE CSE PYQs.

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The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is …

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