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 1

Answer: (D)

$\frac{23}{90}$

Answer 2

Nos. divisible by 6 are 6, 12, 18, ......, 90. Nos. divisible by 8 are 8, 16, 24, ......, 88. Now, total no. divisible by 6 = 15 and total no. divisible by 8 = 11 Now, the no. divisible by both 6 and 8 are 24, 48, 72. $\therefore$ Probability (number divisible by 6 or 8) $= \frac{15 + 11 - 3}{90} = \frac{23}{90}$