Maths M.C.Q

2.  $\frac{29}{40}$

Solution:

Total number of bulbs = 80

Probability that selected bulb has a life more than 500 hours $=\frac{23+25+10}{80}$

$=\frac{58}{80}$

$=\frac{29}{40}$ 

3. $\frac{11}{15}$

Solution:

Total number of outcomes = 600

Probability of getting at least 1 head = Probability of getting 1 head + Probability of getting 2 heads

$=\frac{206}{600}+\frac{234}{600}$

$=\frac{440}{600}$

$=\frac{11}{15}$

3. $\frac{3}{4}$

Solution:

Total number of children = 364

Number of children who like to eat potato chips = 91

⇒ Number of children who do not like to eat potato chips = 364 - 91 = 273

$\therefore$ Required probability $=\frac{273}{364}=\frac{3}{4}$

3. $\frac{1}{5}$

Solution:

Total number of students = 11 + 15 + 8 + 6 = 40

Number of students having blood group AB = 8 

$\therefore$ Required probaility $=\frac{8}{40}=\frac{1}{5}$

4. $\frac{1}{12}$

Solution:

Let E be the event of choosing a student born in october.

P(E) $=\frac{\text{Number of students born in october}}{\text{Toatl number of student}}$

$=\frac{3}{36}$

$=\frac{1}{12}$

3. $\frac{4}{5}$

Solution:

Let E be the event where the ball does not hit a boundary.

The batsman hits a boundary 6 times.

So, he does not hit a boundary 30 - 6 = 24 times

P(E) $=\frac{\text{Number of times the ball does not hit a boundary}}{\text{Total number of balls played}}$

$=\frac{24}{30}$

$=\frac{4}{5}$ 

4. 0

Solution:

Total number of bulbs = 80

Number of bulbs having life of 1150 hours = 0

$\therefore$ Required probability = 0

4. $\frac{4}{5}$

Solution:

Total number of people = 645

Number of people having high school certificate = 516

$\therefore$ Required probability $=\frac{516}{645}=\frac{4}{5}$

3. $\frac{13}{40}$

Solution:

Total number of students = 200

Number of students who does not like Sanskrit = 65

$\therefore$ Required probability $=\frac{65}{200}=\frac{13}{40}$

2. $\frac{4}{5}$

Solution:

Total number of outcomes = 1000

Probability of getting at most 1 head = Probability of getting at most 0 head + Probability of getting at 1 head

$=\frac{250}{1000}+\frac{550}{1000}$

$=\frac{800}{1000}$

$=\frac{4}{5}$

3.  $\frac{5}{12}$

Solution:

Total number of times the coin is tossed = 60

Number of times tail appears = 35

So, the number of times head appears = 60 - 35 = 25

Required probability $=\frac{\text{Number of times head appears}}{\text{Total number of rimes the coin is tossed}}$

$=\frac{25}{60}$

$=\frac{5}{12}$ 

2. $\frac{2}{5}$

Solution:

Total number of balls in the bag = 5 + 8 + 7 = 20

Number of black balls = 8

Required probability $=\frac{\text{Number of black balls}}{\text{Total number of balls}}$

$=\frac{8}{20}$

$=\frac{2}{5}$

1.  $\frac{12}{25}$

Solution:

Let E be the event of getting an evan number.

So, E = {2, 4, 6}

2 appears 9 times, 4 appears 7 times and 6 appears 8 times.

So, the total number of times an even number comes up is 9 + 7 + 8 = 24

P(E) $=\frac{\text{Number of times an even number comes up}}{\text{Total number of times the die is thrown}}$

$=\frac{24}{50}$

$=\frac{12}{25}$  

2. $\frac{5}{16}$

Solution:

Total number of cards in the bag = 16

Number divisible by 3 are {3, 6, 9, 12, 15}

Number of card that bears numbers divisible by 3 = 5

Required probability $=\frac{5}{16}$

3. $\frac{31}{65}$

Solution:

Let E be the event of getting a prime number.

So, E = {2, 3, 5}

2 appears 10 times, 3appears 12 times and 5 appears 9 times.

So, the total number of times a prime number comes up is 10 + 12 + 9 = 31.

P(E) $=\frac{\text{Number of times a prime number comes up}}{\text{Total number of times the die is thrown}}$

$=\frac{31}{65}$