Question

A box contains 19 balls bearing numbers 1, 2, 3,..., 19 respectively. A ball is drawn at random from the box. Find the probability that the number on the ball is

(i) a prime number

(ii) divisible by 3 or 5

(iii) neither divisible by 5 nor by 10

(iv) an even number.

​

Answer 1

☯ AnSwEr :

(i) A prime number.

→ Favourable outcome = 8 (because there are 8 prime numbers i.e., 2, 3, 5, 7, 11, 13, 17, 19)

→ Total outcomes = 19 (because there are total 19 balls).

We know that,

$\large{\star{\boxed{\sf{Probability = \dfrac{Favourable \: outcomes}{Total \: outcomes} }}}}$

★ Putting Values ★

$\sf{\dashrightarrow Probability = \dfrac{8}{19}} \\ \\ \Large{\implies{\boxed{\boxed{\sf{Probability = \dfrac{8}{19} }}}}}$

$\rule{200}{2}$

(ii) Divisible by 3 or 5

→ Favourable outcome = 8 (because there are 8 numbers which are divisible by 3 and 5 i.e., 3, 5, 6, 9, 10, 12, 15, 18)

→ Total outcomes = 19 (because there are total 19 balls).

We know that,

$\large{\star{\boxed{\sf{Probability = \dfrac{Favourable \: outcomes}{Total \: outcomes} }}}}$

★ Putting Values ★

$\sf{\dashrightarrow Probability = \dfrac{8}{19}} \\ \\ \Large{\implies{\boxed{\boxed{\sf{Probability = \dfrac{8}{19} }}}}}$

$\rule{200}{2}$

(iii) Neither divisible by 5 nor by 10.

→ Favourable outcome = 11 (bexause there are 11 numbers which are neither divisible by 5 nor by 10 I.e., 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19.)

→ Total outcomes = 19 (because there are total 19 balls).

We know that,

$\large{\star{\boxed{\sf{Probability = \dfrac{Favourable \: outcomes}{Total \: outcomes} }}}}$

★ Putting Values ★

$\sf{\dashrightarrow Probability = \dfrac{16}{19}} \\ \\ \Large{\implies{\boxed{\boxed{\sf{Probability = \dfrac{16}{19} }}}}}$

$\rule{200}{2}$

(iv) An even number

→ Favourable outcome = 9 (because there are 9even numbers i.e., 2, 4, 6, 8, 10, 12, 14, 16, 18)

→ Total outcomes = 19 (because there are total 19 balls).

We know that,

$\large{\star{\boxed{\sf{Probability = \dfrac{Favourable \: outcomes}{Total \: outcomes} }}}}$

★ Putting Values ★

$\sf{\dashrightarrow Probability = \dfrac{9}{19}} \\ \\ \Large{\implies{\boxed{\boxed{\sf{Probability = \dfrac{9}{19} }}}}}$

$\rule{200}{2}$

Answer 2

Answer:

↪↪↪↪

(i) 8/19 (ii) 8/19 (iii) 11/19 (iv) 9/19

step by step explanation➡

$\huge\mathbb{ \red{SOLUTION}} \\ \\ \sf\ (i) a \: prime \: number \\ \\ \sf\ Favourable \: outcome = 8 \\ \bf\ (\because \:there \: are \: 8 \: prime \: numbers \to: 2, 3, 5, 7, 11, 13, 17, 19) \\ \\ \sf\ Total \: outcomes = 19 \\ \sf\ ( \because \: there \: are \: total \: 19 \: balls). \\ \\ \bf\red{using \: formula \: probabiliy \implies: \frac{Favourable \: outcome }{Total \: outcomes} } \\ \\ \sf\ \implies:probabiliy = \frac{8}{19} \\ \\ \\ \sf\ (ii) divisible \: by \: 3 \: or \: 5</p><p> \\ \\ \sf\ Favourable \: outcome = </p><p>8 \\ \bf\ \: ( \because \: there \: are \: 8 \: numbers \: which \: are \: divisible \: by \: 3 \: and \: 5 \to: 3, 5, 6, 9, 10, 12, 15, 18) \\ \\ </p><p>\sf\ Total \: outcomes = 19 \\ \sf\ ( \because \: there \: are \: total \: 19 \: balls). \\ \\ \bf\red{using \: formula \: probabiliy \implies: \frac{Favourable \: outcome }{Total \: outcomes} } \\ \\ \sf\ \implies:probabiliy = \frac{8}{19} \\ \\ \sf\ (iii) Neither \: divisible \: by \: 5 \: nor \: by \: 10. \\ \\ </p><p></p><p> \\ \sf\ Favourable \: outcome = 11 \\ \bf\ ( \because \:there \: are \: 11 \: numbers \: which \: are \: neither \: divisible \: by \: 5 \: nor \: by \: 10 \to: \\ \bf\ 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19.) \\ \\ \sf\ Total \: outcomes = 19 (because \: there \: are \: total \: 19 \: balls). \\ \\ \bf\red{using \: formula \: probabiliy \implies: \frac{Favourable \: outcome }{Total \: outcomes} } \\ \\ \sf\ \implies:probabiliy = \frac{16}{19} \\ \\ \bf\ (iv) An even number \\ \\ \sf\ \: Favourable \: outcome = 9 \\ \bf\ (\because \: there \: are \: 9 \: even \: numbers \: \to \:2,4,6,8,10,12,14,16,18) \\ \\ \sf\ Total \: outcomes = 19 (because \: there \: are \: total \: 19 \: balls).</p><p>\bf\red{using \: formula \: probabiliy \implies: \frac{Favourable \: outcome }{Total \: outcomes} } \\ \\ \sf\ \implies:probabiliy = \frac{9}{19} </p><p></p><p>$