From a set of 100 cards numbered 1 to 100, one card is drawn at random. The probability that the number obtained on the card is divisible by 6 or 8 but not by 24, is:

Question

From a set of 100 cards numbered 1 to 100, one card is drawn at random. The probability that the number obtained on the card is divisible by 6 or 8 but not by 24, is: (A)  (6/25)  (B)  (1/4)  (C)  (1/6)  (D)  (2/5)

Tardigrade Admin · Accepted Answer

E = set of number divisible by 6 = 6,12,18, text 24, text 30,...,96 F = set of number divisible by 8 = 8,16, text 24,...,96 E∩ F= set of number divisible by 6 and 8 = 24, text 48, text 72, text 96 ∴ n(E)=16,n(F)=12,n(E∩ F)=4 ∴ n(E∪ F)=n(E)+n(F)+n(F)-n(E∩ F) =16+12-4=24 and n(S)=100. ∴ Required probability (n(E∪ F)/n(S))=(24/100)=(6/25)

Q. From a set of 100 cards numbered 1 to 100, one card is drawn at random. The probability that the number obtained on the card is divisible by 6 or 8 but not by 24, is:

$\frac{6}{25}$

$\frac{1}{4}$

$\frac{1}{6}$

$\frac{2}{5}$

$\frac{4}{5}$