Jee Main 2024 29 01 2024 Shift 2 - Question1

Answer: (1)

Solution:

Take $P(A)=$ Probability that number is multiple of 4 $P(B)=$ Probability that number is multiple of 6 $P(C)=$ Probability that number is multiple of 7

$P(A)=\frac{12}{50}, P(B)=\frac{8}{50}, P(C)=\frac{7}{50}$

$P(A \cap B)=\frac{4}{50}$ (Multiple of 12)

$P(B \cap C)=\frac{1}{50}$ (Multiple of 42)

$P(A \cap C)=\frac{1}{50}$ (Multiple of 28)

$P(A \cap B \cap C)=0$ (Multiple of 84)

$P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-$

$P(B \cap C)-P(A \cap C)+P(A \cap B \cap C)$

$=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}+0$

$=\frac{21}{50}$