Question

8.
A bag contains 6 white balls numbered from 1 to 6 and 4 red balls numbered from
7 to 10. A ball is drawn at random. Find the probability of getting
i) an even-numbered ball
(ii) a white ball.​

Answer 1

Answer:

i) 5/10 or 1/2

ii) 6/10 or 3/5

Answer 2

⚘GIVEN

• Total no. of white balls marked as 1 to 6 = 6
• Total no. of red balls marked as 7 to 10 = 4

⚘FIND

• Probability of getting drawn an even-numbered ball
• Probability of getting an White ball

⚘SOLUTION

i) An even-numbered ball

Total numbers of Balls marked as even numbers are = 2,4,6,8,10 = 5

Total number of balls = No. of white balls + No. of red balls= 6 + 4 = 10

We know, that

$\rm \to Probability \: of \: an \: Event = \frac{favourable \: outcomes}{total \: outcomes}$

where,

• Favourable outcomes = 5
• Total outcomes = 10

So,

$\rm \to Probability ( E_{even \: no.}) = \frac{5}{10}$

$\rm \to Probability ( E_{even \: no.}) = \frac{ \cancel{5}}{ \cancel{10} }$

$\rm \to Probability ( E_{even \: no.}) = \frac{1}{2}$

Hence, Probability of getting an even numbered ball is $\rm \frac{1}{2}$

ii) A white ball

Total numbers of white balls = 6

Total number of balls = No. of white balls + No. of red balls= 6 + 4 = 10

we, know that

$\rm \to Probability \: of \: an \: Event = \frac{favourable \: outcomes}{total \: outcomes}$

where,

• Favourable outcomes = 6
• Total outcomes = 10

So,

$\rm \to Probability ( E_{white \: ball}) = \frac{6}{10}$

$\rm \to Probability ( E_{white \: ball}) = \frac{ \cancel{6}}{ \cancel{10} }$

$\rm \to Probability ( E_{white \: ball}) = \frac{3}{5}$

Hence, Probability of getting an White coloured ball is $\rm \frac{3}{5}$