Question

find the probability of choosing an even number from the first 7 even natural numbers​

Answer 1

$\large\bold\orange{\underline{Solution :-}}$

All possible outcomes are:- 1, 2, 3, 4, 5, 6 and 7

• Total number of possible outcomes = 7

Let E be the event of getting a even natural numbers.

• Favourable outcomes are:- 2, 4 and 6
• Number of favourable outcomes = 3

We know that,

$P(E) = \dfrac{Number \: of \: favorable \: outcomes}{Total \: number \: of \: possible \: outcomes}$

$=\dfrac{3}{7}= </p><p>7</p><p>3</p><p> </p><p> </p><p>$

$∴ P(E) = \dfrac{3}{7} </p><p>7</p><p>3</p><p></p><p>$

Answer 2

1 is the answer because all the numbers are even as said in the question