Comedic Luck

Probability Level 2

A positive integer is a Nabeatsu number if it is divisible by $3$ , or has $3$ in at least one of its decimal places. For example, while $31$ is not divisible by $3$ , it has $3$ in the tens digit. For this problem,

Let $P_2$ denote the probability of choosing a number divisible by $2$ from the set of positive integers from $1$ to $39$ .
Let $P_3$ denote the probability of choosing a Nabeatsu number from the set of positive integers from $1$ to $39$ .

Which of the following statements must be true?

Assumption: The values of $P_2$ and $P_3$ are between $0$ and $1$ .

P_2 > P_3

P_2 = P_3

P_2 < P_3

1 solution

Agent T
May 22, 2021

Let the set of no.divible by 2 be T and Nabeatsu numbers be Z

T={2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38}

Z={3,6,9,12,13,15,18,21,23,24,27,30,31,32,33,34,35,36,37,38,39}

$P_{2} =\dfrac{19}{39}$

$P_{3} =\dfrac{21}{39}$

Hence $\boxed{P_{3} \textcolor{#D61F06}{\boxed{ > }} P_{2}}$

Boom!

Ik there must be some very short ans to this ques that is yet into the unknown.....

Agent T - 3 weeks ago

Slight error... For $P_3$ , the correct count is $21$ and not $22$ as the number $1$ does not contain decimal digit $3$ .

Michael Huang - 2 weeks, 6 days ago

Okay ,thanks!

Agent T - 2 weeks, 6 days ago

Comedic Luck

1 solution

Boom!

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