What is the probability that a positive two digit number is divisible by 2, 3, and 5?

$a$ and $b$ are $2$-digit positive integers, so $\, a,b \in \{1,2,3,...,99\}$.

There are exactly $33$ numbers in $\{1,2,3,...,99\}$ that are divisible by $3$, namely

$$\{3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99\}$$

The same statement is true for $\; b+1\in\{2,3,4,...,100\}$. $$$$

\begin{align} P\left(3 \;\text{divides} \; ab+a \right)&= P\left(3 \;\text{divides} \; a(b+1)\right) \\ &= P\left(3 \;\text{divides} \; a\;\text{OR}\; 3 \;\text{divides} \; b+1\right) \\ &=P\left(3 \;\text{divides} \; a\right)+P\left(3 \;\text{divides} \; b+1\right)-P\left(3 \;\text{divides} \; a\;\text{AND}\; 3 \;\text{divides} \; b+1\right) \\ &=\frac{33}{99}+\frac{33}{99}-\frac{33}{99}\cdot\frac{33}{99} \\ &=\frac{1}{3}+\frac{1}{3}-\frac{1}{3}\cdot\frac{1}{3} \\ &=\frac{5}{9} \end{align}

Tardigrade - CET NEET JEE Exam App

© 2022 Tardigrade®. All rights reserved

What is the probability that a two digit number selected at random will be a multiple of ' 3 ' and not a multiple of '5'?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App