3 Marks Question

Question 13 Marks

A bag contains $5$ red marbles, $8$ white marbles, $4$ green marbles. What is the probability that if one marble is taken out of the bag at random, it will be
$i.$ red
$ii.$ white
$iii.$ not green

Answer

Number of red marbles $= 5$
Number of white marbles $= 8$
Number of green marbles $= 4$
Total number of marbles in the bag $= 5 + 8 + 4 = 17$
$\therefore$ Total number outcomes $= 17$
$i.$ Let $A$ be the event of drawing a red ball.
$\therefore\text{ P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{5}{17}$
$ii.$ Let $B$ be the event of drawing a white ball.
$\therefore\text{ P(B)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{8}{17}$
$iii.$ Let $C$ be the event of drawing a green ball.
$\therefore\text{ P(C)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{4}{17}$
Now, the event of not drawing a green ball is:
$\text{P}(\bar{\text{C}}) = 1 −\text{ P(C)} = 1 − \frac{4}{17} = \frac{13}{17}$

View full question & answer→

Question 23 Marks

If you have a collection of $6$ pairs of white socks and $3$ pairs of black socks. What is the probability that a pair you pick without looking is
$i.$ white$?$
$ii.$ black$?$

Answer

Number of pairs of white socks $= 6$
Number of pairs of black socks $= 3$
Total number of pairs of socks $= 6 + 3 = 9$
$\therefore$ Number of possible outcomes $= 9$
Let $A$ be the event of getting a pair of white socks.
$ \therefore\text{P(A)}=\frac{6}{9}=\frac{2}{3}$
Let $B$ be the event of getting a pair of black socks.
$ \therefore\text{P( B)}=\frac{3}{9}=\frac{1}{3}$

View full question & answer→

Question 33 Marks

If you have a spinning wheel with $3-$green sectors, $1-$blue sector and $1-$red sector. What is the probability of getting a green sector$?$ Is it the maximum$?$

Answer

Number of green sectors in the wheel$ = 3$
Number of blue sectors in the wheel $= 1$
Number of red sectors in the wheel $= 1$
Total number of sectors in the wheel $= 3 + 1 + 1 = 5$
$\therefore$ Number of possible outcomes $= 5$
$\therefore\text{P(A)}=\frac{3}{5}, \text{P(B)}=\frac{1}{5} \text{and}\text{ P(C)}=\frac{1}{5}$
Hence, the probability of getting a green sector is the maximum.

View full question & answer→

Question 43 Marks

If you put $21$ consonants and $5$ vowels in a bag. What would carry greater probability$?$ Getting a consonant or a vowel$?$ Find each probability.

Answer

Number of consonants $= 21$
Number of vowels $= 5$
Total number of possible outcomes $= 21 + 5 = 26$
Let $C$ be the event of getting a consonant and $V$ be the event of getting a vowel.
$\therefore\text{P(C)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{21}{26}$
$\text{And,} \ \text{P(C)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{5}{26}$
Thus, the consonants have a greater probability.

View full question & answer→

Question 53 Marks

A bag contains $3$ red balls and $5$ black balls. $A$ ball is drawn at random from the bag. What is the probability that the ball drawn is:
$i.$ red
$ii.$ black

Answer

Number of red balls $= 3$
Number of black balls $= 5$
Total number of balls $= 3 + 5 = 8$
Let $A$ be the event of drawing a red ball.
$\therefore\text{ P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{3}{8}$
Let $B$ be the event of drawing a black ball.
$\therefore\text{ P( B)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{5}{8}$

View full question & answer→

Maths 3 Marks Question

Take a timed test