We are given a set of first 100 naturals numbers. If the probability of the product of two numbers selected from this set be divisible by 7 is given by $\frac{p}{q}$ in it's simplest form. Then $\frac{p + q + 1}{p - 9}$ is ___

Open in App

Solution

The probability of product of two numbers divisible by 7 = 1- (Probability of product of two numbers not divisible by 7)

The product of two numbers not divisible by 7 when neither of the two numbers is divisible by 7.

There are 14 numbers from 1 to 100 which are divisible by 7. Therefore there are 86 numbers from 1 to 100 which are not divisible by 7.

Therefore probability $= 1 - \frac{{^{86} C}_{2}}{^{100} C_{2}} = \frac{259}{990}$

Suggest Corrections

Similar questions

Q. If

p

is selected from the set

{1, 2, 3, . . . . . .100}

, then the probability of selecting

p

such that the number

2^{p} + 3^{p} + 5^{p}

is divisible by

4

Q. From a set of first

18

natural number two number

{x, y}

are selected randomly then the probability that sum of their cubes is divisible by

3

, is

Q. Two numbers a and b are selected from the set of natural number then the probability that

a^{2} + b^{2}

is divisible by

5

is?

If three consecutive numbers are selected randomly from the first 100 natural numbers, then what is the probability that their product of the three numbers is divisible by 24?

Q. Two distinct numbers are chosen at random from the set

{1, 2, 3, . . . ., 3 n}

. The probability that

x^{2} - y^{2}

is divisible by

3

\frac{p n + q}{r (3 n - 1)}

, then

p + q + r =

____

Join BYJU'S Learning Program

We are given a set of first 100 naturals numbers. If the probability of the product of two numbers selected from this set be divisible by 7 is given by pq in it's simplest form. Then p+q+1p−9 is ___

We are given a set of first 100 naturals numbers. If the probability of the product of two numbers selected from this set be divisible by 7 is given by $\frac{p}{q}$ in it's simplest form. Then $\frac{p + q + 1}{p - 9}$ is ___