4 Mark Question

Question 14 Marks

A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:

white?
red?
black?
not red?

Answer

Number of red balls = 3
Number of black balls = 5
Number of white balls = 4
Total number of balls = 3 + 5 + 4 = 12
Therefore, the total number of cases is 12.

Since there are 4 white balls, the number of favourable outcomes is 4.

P(a white ball) $=\frac{\text{Number of favourable cases}}{\text{Total number of cases}}=\frac{4}{12}=\frac{1}{3}$

Since there are 3 red balls, the number of favourable outcomes is 3.

P(a red ball) $=\frac{\text{Number of favourable cases}}{\text{Total number of cases}}=\frac{3}{12}=\frac{1}{4}$

Since there are 5 black balls, the number of favourable outcomes is 5.

P(a black ball) $=\frac{\text{Number of favourable cases}}{\text{Total number of cases}}=\frac{5}{12}$

P(not a red ball) $=1−\text{P}\text{(a red ball)}=1−\frac{1}{4}=\frac{3}{4}$

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Question 24 Marks

A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white?

Answer

Number of white balls = 5
Number of red balls = 7
Total number of balls = 5 + 7 = 12
$\therefore$The total number of cases = 12
P(the drawn ball is white) $=\frac{\text{Number of favourable cases}}{\text{Total number of cases} }=\frac{5}{12}$

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Question 34 Marks

If we have 15 boys and 5 girls in a class which carries a higher probability? Getting a copy belonging to a boy or a girl. Can you give it a value?

Answer

Number of boys in the class = 15
Number of girls in the class = 5
Total number of students in the class = 15 + 5 = 20
$\therefore$ Number of possible outcomes = 20
Since the number of boys is more than the number of girls, boys will have a higher probability.
Hence, there is the higher probability of getting a copy belonging to a boy.
Let A be the event of getting a boy's copy and B be the event of getting a girl's copy.
$\therefore\text{P(A)} =\frac{15}{20}=\frac{3}{4} $
$\text{And}, \text{P}\text{(B)} =\frac{5}{20}=\frac{1}{4}$

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Question 44 Marks

A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

white
red
not black
red or white

Answer

Number of red balls = 4
Number of black balls = 5
Number of white balls = 6
Total number of balls in the bag = 4 + 5 + 6 = 15
Therefore, the total number of cases is 15.

Let A denote the event of getting a white ball.

Number of favourable outcomes, i.e. number of white balls = 6

$\text{P(A)} =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{6}{15}=\frac{2}{5}$

Let B denote the event of getting a red ball.

Number of favourable outcomes, i.e. number of red balls = 4

$\text{P(B)} =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{4}{15}$

Let C denote the event of getting a black ball.

Number of favourable outcomes, i.e. number of black balls = 5

$\text{P(C)} =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{5}{15}=\frac{1}{3}$

Therefore, the probabilty of not getting a black ball is as follows:

$\text{P}(\bar{\text{C}}) = 1 − \text{P(C)} = 1 − \frac{1}{3} =\frac{ 2}{3}$

Let D denote the event of getting a red or a white ball.

$\text{P(D)} =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\=\frac{4 + 6}{15}=\frac{10}{15}=\frac{2}{5}$

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