Selecting-randomly-an-integer-from-1-to-2000-find-the-probability-that-neither-6-nor-8-divides-the-integer-

Question Number 131631 by liberty last updated on 07/Feb/21

Selecting randomly an integer

from 1 to 2000, find the probability

that neither 6 nor 8 divides the

integer .

Answered by som(math1967) last updated on 07/Feb/21

p r o b a b i l i t y d i v i s i b l e b y 6

\frac{333}{2000}

[I n t h e s e r e i e s 6, 12 \dots . 1998

1998 = 6 + (n - 1) \times 6 ∴ n = 333]

p r o b a b i l i t y d i v i s i b l e b y 8

= \frac{250}{2000} [8 + (n - 1) \times 8 = 2000

∴ n = 250]

p r o b a b i l i t y d i v i s i b l e b y b o t h

6, 8 = \frac{83}{2000}

[1992 = 24 + (n - 1) \times 24

n = 83]

∴ p r o b a b i l i t y d i v i s i b l e b y

6 o r 8 = \frac{333}{2000} + \frac{250}{2000} - \frac{83}{2000}

= \frac{500}{2000}

∴ n e i t h e r 6 n o r 8

1 - \frac{500}{2000} = \frac{1500}{2000} = \frac{3}{4}

Commented by liberty last updated on 07/Feb/21

typo sir it should be \frac{250}{2000} [8 n] = 2000

n = 250

Commented by som(math1967) last updated on 07/Feb/21

y e s s i r, t h a n k y o u