5 Marks Questions

Question 15 Marks

A die is thrown. Find the probability of getting:
$i.$ A prime number.
$ii. 2$ or $4.$
$iii.$ A multiple of $2$ or $3.$

Answer

When a die is thrown, the possible outcomes are $1, 2, 3, 4, 5$ and $6$.
Thus, the sample space will be as follows: $S = \{1, 2, 3, 4, 5, 6\}$
$i.$ Let $A$ be the event of getting a prime number.
There are $3$ prime numbers $(2, 3$ and $5$) in the sample space.
Thus, the number of favourable outcomes is $3.$
Hence, the probability of getting a prime number is as follows:
$\text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{3}{6}=\frac{1}{2}$
$ii.$ Let $A$ be the event of getting a two or four.
Two or four occurs once in a single roll.
Therefore, the total number of favourable outcomes is $2.$
Hence, the probability of getting $2$ or $4$ is as follow:
$\text{P(A)}=\frac{2}{6}=\frac{1}{3}$
$iii.$ Let $A$ be the event of getting multiples of $2$ or $3.$
Here, the multiples of $2$ are $2, 4, 6$ and the multiples of $3$ are $3$ and $6.$
Therefore, the favourable outcomes are $2, 3, 4$ and $6.$
Hence, the probability of getting a multiple of $2$ or $3$ is as follows:
$\text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{4}{6}=\frac{2}{3}$

View full question & answer→

Question 25 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:
$i.$ white?
$ii.$ red?
$iii.$ black?
$iv.$ 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 $12.$
$i.$ Since there are $4$ white balls, the number of favourable outcomes is $4.$
$P($ a white ball$) =\frac{\text{Number of favourable cases}}{\text{Total number of cases}}=\frac{4}{12}=\frac{1}{3}$
$ii.$ Since there are $3$ red balls, the number of favourable outcomes is $3.$
$P($a red ball$) =\frac{\text{Number of favourable cases}}{\text{Total number of cases}}=\frac{3}{12}=\frac{1}{4}$
$iii.$ Since there are $5$ black balls, the number of favourable outcomes is $5.$
$P($a black ball$) =\frac{\text{Number of favourable cases}}{\text{Total number of cases}}=\frac{5}{12}$
$iv.\ P($not a red ball$) =1−\text{P}\text{(a red ball)}=1−\frac{1}{4}=\frac{3}{4}$

View full question & answer→

Question 35 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)
$=\frac{\text{Number of favourable cases}}{\text{Total number of cases} }=\frac{5}{12}$

View full question & answer→

Question 45 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} $
$\text{And}, \text{P}\text{(B)} =\frac{5}{20}=\frac{1}{4}$

View full question & answer→

Question 55 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:
$i.$ white
$ii.$ red
$iii.$ not black
$iv.$ 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 $15.$
$i.$ Let $A$ denote the event of getting a white ball.
Number of favourable outcomes, i.e. number of white balls $= 6$
$\text{P(A)} =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{6}{15}=\frac{2}{5}$
$ii.$ Let $B$ denote the event of getting a red ball.
Number of favourable outcomes, i.e. number of red balls $= 4$
$\text{P(B)} =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{4}{15}$
$iii.$ 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:
$\text{P}(\bar{\text{C}}) = 1 − \text{P(C)} = 1 − \frac{1}{3} =\frac{ 2}{3}$
$iv.$ Let $D$ denote the event of getting a red or a white ball.
$\text{P(D)} =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\=\frac{4 + 6}{15}=\frac{10}{15}=\frac{2}{5}$

View full question & answer→

MATHS 5 Marks Questions

Test yourself on this topic