Question

find the probility of choosing a prime number from the series of natural number 170 ​

Answer 1

Answer:

Theory :

A sum of two perfect cubes, a3 + b3 can be factored into :

\sf\green{ Mass \: of \: the \: object, \:}Massoftheobject,

━━━━━━━━━━━━━━━━━━━━━━━━━

━━━━━━━━━━━━━━━━━━━━━━━━━

Proof :

(a+b) • (a2-ab+b2) =

a3-a2b+ab2+ba2-b2a+b3 =

a3+(a2b-ba2)+(ab2-b2a)+b3=

a3+0+0+b3=

\sf\color{blue}{→a3+b3}→a3+b3

━━━━━━━━━━━━━━━━━━━━━━━━━

━━━━━━━━━━━━━━━━━━━━━━━━━

Check :

27 is the cube of 3

Check :

r3 is the cube of r1

Check :

p3 is the cube of p1

━━━━━━━━━━━━━━━━━━━━━━━━━

━━━━━━━━━━━━━━━━━━━━━━━━━

Factorization is :

\mathrm{\boxed{\boxed{\pink{→ (r + 3p) • (r2 - 3rp + 9p2) ✔}}}}→(r+3p)•(r2−3rp+9p2)✔

Answer 2

Step-by-step explanation:

We have the following numbers that can be picked.

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17

So, there are a total of 17 ways of choosing one number.

We have to find the favourable outcomes.

Since 1 is neither prime, nor composite, we exclude 1 and then find all the primes from 2 to 17 inclusive. These are 2, 3, 5, 7, 11, 13, 17. ( a total of seven numbers are prime)

So, the required probability ( probability of a prime number when one number is selected at random from natural numbers from 1 to 17) = 7/17.