Aptitude Discussion

Q. |
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 |

**Solution:**

Option(**B**) is correct

There are a total of 90 two digit numbers. Every third number will be divisible by '3'. Therefore, there are 30 of those numbers that are divisible by '3'.

Of these 30 numbers, the numbers that are divisible by '5' are those that are multiples of '15'. i.e. numbers that are divisible by both '3' and '5'.

There are 6 such numbers -- 15,30,45,60,75 and 90.

We need to find out numbers that are divisible by '3' and not by '5', which will be:

$30−6=24$.

24 out of the 90 numbers are divisible by '3' and not by '5'.

The required probability is therefore,

\(=\dfrac{24}{90}\)

\(=\dfrac{4}{15}\)

**Shishir**

*()
*

**Lulu**

*()
*

$P(A) = \dfrac{30}{90} = \dfrac{1}{3}$ -> multiple of 3

$P(B) = \dfrac{6}{90} = \dfrac{1}{15}$ -> multiple of 5

$P(\text{A not B}) = P(A) - P(A \cap B)$

$P(\text{A not B}) = \dfrac{1}{3} - \dfrac{1}{45} = \dfrac{14}{45}$

Which is not the correct answer.

What am I doing wrong?

You are using this formula,

$P(\text{A not B}) = P(A) - P(A \cap B)$

This formula is not applicable here.

This is applicable only in case of mutually exclusive events.

**Sanjeev Raj**

*()
*

I'm sorry. I forgot to count nos. between $90-99.$

**Sanjeev Raj**

*()
*

There is a total of 90 (not two digits) numbers which are multiple of 3(it includes 3, 6, & 9 also). Every third number will be divisible by '3'. Therefore, there are 30 of those numbers that are divisible by '3'.

So, total 27 two digit numbers are multiple of 3.

Of these 27 numbers, the numbers that are divisible by '5' are those that are multiples of '15'. i.e. numbers that are divisible by both '3' and '5'.

There are 6 such numbers -- 15,30,45,60,75 and 90.

We need to find out numbers that are divisible by '3' and not by '5', which will be:

$27-6=21.$

21 out of the 90 numbers are divisible by '3' and not by '5'.

The required probability is therefore,

$=\dfrac{21}{90}$

$= \dfrac{7}{30}$

I guess you are making a mistake in the first line of your solution, 'There is a total of 90 (not two digits) numbers which are multiple of 3(it includes 3, 6, & 9 also).'

There will be 33 and not 30 such numbers. so in order to get the correct answer '3' should be added to your favorable cases. Which changes the final answer to be,

$=\dfrac{21+3}{90}$

$=\dfrac{24}{90}$

$=\dfrac{4}{15}$

There are a total of 91 two digit numbers. (10 is a two digit number)