A ticket is drawn from 200 tickets numbered f | vedclass

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Question

(A) Let $A,$ $B,$ and $C$ be the events that the number is divisible by $2,$ $3,$ and $5$ respectively.Total number of tickets $n(S) = 200.$$n(A) = \l

Vedclass · Accepted Answer

$\frac{73}{100}$

Vedclass · Answer

(A) Let $A,$ $B,$ and $C$ be the events that the number is divisible by $2,$ $3,$ and $5$ respectively.Total number of tickets $n(S) = 200.$$n(A) = \lfloor \frac{200}{2} \rfloor = 100, n(B) = \lfloor \frac{200}{3} \rfloor = 66, n(C) = \lfloor \frac{200}{5} \rfloor = 40.$$n(A \cap B) = \lfloor \frac{200}{LCM(2,3)} \rfloor = \lfloor \frac{200}{6} \rfloor = 33.$$n(B \cap C) = \lfloor \frac{200}{LCM(3,5)} \rfloor = \lfloor \frac{200}{15} \rfloor = 13.$$n(C \cap A) = \lfloor \frac{200}{LCM(5,2)} \rfloor = \lfloor \frac{200}{10} \rfloor = 20.$$n(A \cap B \cap C) = \lfloor \frac{200}{LCM(2,3,5)} \rfloor = \lfloor \frac{200}{30} \rfloor = 6.$Using the Principle of Inclusion-Exclusion:$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$$n(A \cup B \cup C) = 100 + 66 + 40 - 33 - 13 - 20 + 6 = 146.$Required probability $= \frac{n(A \cup B \cup C)}{n(S)} = \frac{146}{200} = \frac{73}{100}.$

$A$ ticket is drawn from $200$ tickets numbered from $1$ to $200.$ Find the probability that the number is divisible by $2,$ $3,$ or $5.$

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