Question

a bag containing 20 balls marked 1 to 20. One ball is drawn at random from the bag. What is the probability that the ball drawn is marked with a number which is a) a multiple of 5? b) a multiple of 7? c) a multiple of 5 or 7?

Answer 1

Given:

• A bag is containing 20 balls marked 1 to 20.
• One ball is drawn at random from the bag.

To Find:

The probability that the ball drawn is marked with a number which is:

a) a multiple of 5

b) a multiple of 7

c) a multiple of 5 or 7

Solution:

We know that,

If a random experiment is conducted, then

P(occurrence of event A)= $\sf{\dfrac{n(E)}{n(S)}}$

where,

• n(E) is total no. of favourable outcomes of A

• n(S) is total no. possible outcomes in an experiment

Here,

Total no. of possible outcomes,n(S)= 20

[Since, no. of balls are 20 and only one ball is drawn at a time]

$\rule{200}{2}$

a) Let E₁ be the event of getting a ball marked with multiple of 5

E₁= {5,10,15,20}

n(E₁)= 4

$\longrightarrow\sf{P(E_{1})=\dfrac{n(E_{1})}{n(S)}}$

$\longrightarrow\sf{P(E_{1})=\dfrac{4}{20}}$

$\longrightarrow\sf{P(E_{1})=\dfrac{1}{5}}$

$\rule{200}{2}$

b) Let E₂ be the event of getting a ball marked with multiple of 7

E₂= {7,14}

n(E₂)= 2

$\longrightarrow\sf{P(E_{2})=\dfrac{n(E_{2})}{n(S)}}$

$\longrightarrow\sf{P(E_{2})=\dfrac{2}{20}}$

$\longrightarrow\sf{P(E_{2})=\dfrac{1}{10}}$

$\rule{200}{2}$

c) Let E₃ be the event of getting a ball marked with multiple of 5 or 7

[It means outcome can either be multiple of 5 or multiple of 7]

E₃= {5,7,10,14,15,20}

n(E₃)= 6

$\longrightarrow\sf{P(E_{3})=\dfrac{n(E_{3})}{n(S)}}$

$\longrightarrow\sf{P(E_{3})=\dfrac{6}{20}}$

$\longrightarrow\sf{P(E_{3})=\dfrac{3}{10}}$

$\rule{200}{2}$

Hence, required probabilities are $\dfrac{1}{5},\dfrac{1}{10}and \dfrac{3}{10}$.