MATHS M.C.Q. [1 Marks Each]

Given, head appeared $120$ times, if a coin is tossed $200$ times. Then,
probability of getting a head in this experiment $=\frac{\text{Number of times head appeared}}{\text{Number of times coin is tossed}}$
$=\frac{120}{200}$
$=\frac{3}{5}$

Impossible event is an event that is non existing. There is no possibility that it can happen. Hence the probability of occurrence is $0.$

When two coins are tossed, then the possible outcomes will be:
$HH, HT, TH, TT$
Thus, probability of getting tails for both the coins $=\frac{1}{4}$

Two coins are tossed simultaneously, then possible outcomes $= 4$
Now probability of getting one head and one tail $=\frac{2}{4}=\frac{1}{2}$

Explanation: For any class, we can determine the class interval, by subtracting the lower limit from upper limit.
$500 - 450 = 50$
$450 - 400 = 50$

Highest frequency $= 7 \rightarrow 430-465$

$8$ cards are numbered $1, 2, 3, 4, 5, 6, 7, 8$
They are mixed and kept in a box one care is chosen at random, then probability of card having 9 number less than $4=\frac{3}{8}$

Total number of cards $= 5$
Number of cards having $2$ on it $= 2$
Probability of Rohan picking a card without seeing, that has the number $2$ on it $\frac{2}{5}$

Size $= 325 - 290 = 35$

Size $= 20 - 15 = 5$

Since the outcome of a throw of dice can never be $10$, the probability is $0.$

Total number of times coin tossed $= 12$
Total number of occurrence of head $= 5$
The chance of occurrence of head $=\frac{\text{Number of times headappeared}}{\text{Number of times a coinis tossed}}$
$=\frac{5}{12}$

Numbers on the slips are $1, 2, 3, 4$ and $5.$
Odd numbers $= 1, 3, 5$
Number of slips bears an odd number $= 3$
Probability that the slip bears an odd number $=\frac{\text{Number of slipsbearsan odd number}}{\text{Total number of slips}}$
$=\frac{3}{5}$

Required number
$=$ Frequency of class $(4 - 6) = 4$

Size of the class $150-175$
= Upper limit - Lower
Limit $= 175 - 150 = 25$

Red $\rightarrow 45\%$
Yellow $\rightarrow 45\%$

A die is thrown possible outcomes $= 6$
Now probability of getting an even number which are $2, 4, 6 =\frac{3}{6}=\frac{1}{2}$

Number of employees who joined company $A$ from january to april
$= 600 + 500 + 400 + 300 = 1800$
Number of employees who joined company $B$ from january to april
$= 400 + 200 + 100 + 100 = 800$
In company A there are $1000$ people more than company $B.$

Population of $USA$ in the year $1981$ and $1991$ together $= 45,000 + 60,000 = 105,000$
Population of $USA$ in the year $1996$ and $2001$ together $= 75,000 + 105,000 = 180,000$
The difference in the population of $USA$ in the year $1981$ and $1991$ together and in the year $1996$ and $2001$ together$= 180,000 - 105,000 = 75,000$

Lowest frequency $= 2 \rightarrow 325-360$

Central angle for sector $B$
$=\frac{40}{100}\times 360^\circ = 144^\circ$

Minimum number of students
$= 2 \rightarrow 8-10$

As, jar contains $6$ red, $5$ green, $4$ blue and $5$ yellow marbles of same size. Then,
Probability that the chosen marble is of red colour $=\frac{\text{Number of red marbles}}{\text{Total number of marbles}}$
$=\frac{6}{6+5+4+5}$
$=\frac{6}{20}$
$=\frac{3}{10}$

Maximum number of students$= 9 \rightarrow 6 – 8$

The number of student having height more than equal to $130\ cm$ but less than $145\ cm= 8 + 14 + 18 = 40$