Question

A number is selected from first 30 natural numbers.What is the probability that
if would be divisible by 2 or it is a prime​

Answer 1

• GIVEN:-

A number is selected from first 30 natural numbers.

• To Find:-

Probability that if would be divisible by 2 or it is a prime.

• SOLUTION:-

We know that,

$\large\boxed{\sf{\purple{Probability=\frac{No \:of \:favourable\: outcomes}{Total \:no \:of\: possible\: outcomes}}}}$

Possible outcomes = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,14 ,15, 16, 17, 18, 19, 20,21, 22, 23, 24, 25, 26, 27, 28, 29, 30.

____________________________

Favourable outcomes for divisible by 2 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30.

Total outcomes = 15

So,

Probability for divisible by 2,

$\large\Longrightarrow{\sf{P(Divisible)=\frac{No \:of \:favourable\: outcomes}{Total \:no \:of\: possible\: outcomes}}}$

$\large\Longrightarrow{\sf{P(Divisible)=\frac{15}{30}}}$

$\large\green\therefore\boxed{\sf{\green{P(Divisible)=\frac{1}{2}}}}$

____________________________

NOW,

Favourable outcomes for prime number = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Total outcomes = 10.

$\large\Longrightarrow{\sf{P(Prime)=\frac{No \:of \:favourable\: outcomes}{Total \:no \:of\: possible\: outcomes}}}$

$\large\Longrightarrow{\sf{P(Prime)=\frac{10}{30}}}$

$\large\red\therefore\boxed{\sf{\red{P(Prime)=\frac{1}{3}}}}$

____________________________

Now by adding P(Divisible) + P(Prime),

$\large\Longrightarrow{\sf{P(Divisible)+P(Prime)=\frac{1}{2}+\frac{1}{3}}}$

$\large\Longrightarrow{\sf{P(Divisible)+P(Prime)=\frac{2+3}{6}}}$

$\large\Longrightarrow{\sf{P(Divisible)+P(Prime)=\frac{5}{6}}}$

$\large\pink\therefore\boxed{\sf{\pink{P(Divisible)+P(Prime)=\frac{5}{6}}}}$

SO,

$\large\blue\therefore\boxed{\sf{\blue{Probability \:of\: getting\: divisible\: by \:2\: or\: prime\: is\:\: \huge\frac{5}{6}}}}$

Answer 2

Answer:

Step-by-step explanation:

let A be the eventof getting a num ber divisible by 2