Question
The probability of choosing at random a number that is divisible by $$6$$ or $$8$$ from among $$1$$ to $$90$$ is equal to :
A.
$$\frac{1}{6}$$
B.
$$\frac{1}{{30}}$$
C.
$$\frac{{11}}{{80}}$$
D.
$$\frac{{23}}{{90}}$$
Answer :
$$\frac{{23}}{{90}}$$
Solution :
Numbers divisible by $$6$$ are $$6,\,12,\,18,\,......,\,90.$$
Numbers divisible by $$8$$ are $$8,\,16,\,24,\,......,\,88.$$
Now, total number divisible by $$6 = 15$$
and total number divisible by $$8 = 11$$
Now, the number divisible by both $$6$$ and $$8$$ are $$24,\,48,\,72.$$
So, total number divisible by both $$6$$ and $$8 = 3$$
$$\therefore $$ Probability ( number divisible by $$6$$ or $$8$$ )
$$ = \frac{{15 + 11 - 3}}{{90}} = \frac{{23}}{{90}}$$