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NCERT Exemplar · Q3

Q.Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

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✓ Free question

By inclusion–exclusion the count is 500+111−55=556500+111-55=556, so P=5561000=139250P=\dfrac{556}{1000}=\dfrac{139}{250}.

Solution

Let AA be the set of multiples of 22 and BB the set of multiples of 99 among {1,2,…,1000}\{1,2,\dots,1000\}. We want P(A∪B)P(A\cup B).

Count each set:

∣A∣=⌊10002⌋=500,∣B∣=⌊10009⌋=111.|A|=\left\lfloor\frac{1000}{2}\right\rfloor=500,\qquad |B|=\left\lfloor\frac{1000}{9}\right\rfloor=111.

A number divisible by both 22 and 99 is divisible by lcm(2,9)=18\mathrm{lcm}(2,9)=18:

∣A∩B∣=⌊100018⌋=55.|A\cap B|=\left\lfloor\frac{1000}{18}\right\rfloor=55.

By the inclusion–exclusion principle,

∣A∪B∣=∣A∣+∣B∣−∣A∩B∣=500+111−55=556.|A\cup B|=|A|+|B|-|A\cap B|=500+111-55=556.

With 10001000 equally likely integers,

P(A∪B)=5561000=139250.P(A\cup B)=\frac{556}{1000}=\frac{139}{250}.

✓Final answer

The probability that the chosen integer is a multiple of 22 or of 99 is 139250=0.556\dfrac{139}{250}=0.556.

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