Can you solve it!

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

A- 2/15
B- 4/15
C- 1/15
D- 4/90

E- 8/11

#Combinatorics

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

There are 30 2-digit numbers 12, 15, 18, , . . . . , 99 that are divisible by 3. Out of these 6 numbers, eg. 15, 30, 45. . . . , 90 are divisible by 5 as well. Hence remaining 24 numbers are favorable cases out of a total of 90 (from 10 to 99). Hence the required probability is 24/90 = 4/15 i.e.B..

Rajen Kapur - 5 years, 8 months ago

I think you missed one digit.

Mehdi Balti - 5 years, 8 months ago

Thanks Mehdi. I have corrected my mistake.

Rajen Kapur - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$