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Mathematics [3 marks sum]

Probability · ICSE STD 10 (ICSE - English Medium). 25 questions.

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[3 marks sum]

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25 questions · timed · auto-graded

Question 13 Marks
An unbiased cubic die is thrown .Find the probability of getting
(iv) a number less than 4.
Answer
When a die is tossed once, the possible outcomes are the numbers $1,2,3,4,5,6$. So, total number of possible outcomes $=6$
The event is getting a number less than 4 and numbers less than 4 are 1,2,3. So, the number of favourable outcomes to the event getting a number less than $4=3$
Therefore, $\mathrm{P}$ (getting a number less than 4 ) =
$
\frac{\text { Favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{6}=\frac{1}{2}
$
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Question 23 Marks
An unbiased cubic die is thrown .Find the probability of getting
(iii) a number divisible by 2
Answer
When a die is tossed once, the possible outcomes are the numbers $1,2,3,4,5,6$. So, total number of possible outcomes $=6$
The event is getting a number divisible by 2 and numbers divisible by 2 are 2,4,6. So, the number of favourable outcomes to the event getting a number divisible by $2=3$
Therefore, $\mathrm{P}($ getting a number divisible by 2$)=\frac{\text { Favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{6}=\frac{1}{2}$
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Question 33 Marks
An unbiased cubic die is thrown .Find the probability of getting
(ii) a number greater than 2
Answer
When a die is tossed once, the possible outcomes are the numbers $1,2,3,4,5,6$. So, total number of possible outcomes $=6$
The event is getting a number greater than 2 and numbers greater than 2 are $3,4,5,6$. So, the number of favourable outcomes to the event getting a number greater than $2=4$
Therefore, P(getting a number greater than 2) $=\frac{\text { Favourable outcomes }}{\text { Total number of outcomes }}=\frac{4}{6}=\frac{2}{3}$
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Question 43 Marks
An unbiased cubic die is thrown .Find the probability of getting (i) an odd number,
Answer
When a die is tossed once, the possible outcomes are the numbers $1,2,3,4,5,6$. So, total number of possible outcomes $=6$
The event is getting an odd number and the odd numbers are $1,3,5$. So, the number of favourable outcomes to the event getting an even number $=3$
Therefore, $\mathrm{P}$ (getting an odd number) $=\frac{\text { Favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{6}=\frac{1}{2}$
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Question 53 Marks
A perfect cubic die is thrown .Find the probability that
(ii) a perfect square comes up.
Answer
When a die is tossed once, the possible outcomes are the numbers $1,2,3,4,5,6$ So, total number of possible outcomes $=6$
The event is getting a perfect square and the perfect squares are 1 and 4 . So, the number of favourable outcomes to the event getting a perfect square $=2$ Therefore, $\mathrm{P}$ (getting a perfect square) $=\frac{\text { Favourable outcomes }}{\text { Total number of outcomes }}=\frac{2}{6}=\frac{1}{3}$
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Question 63 Marks
A perfect cubic die is thrown .Find the probability that
(i) an even number comes up
Answer
When a die is tossed once, the possible outcomes are the numbers $1,2,3,4,5,6$
So, total number of possible outcomes $=6$
(i)The event is getting an even number and the even numbers are 2,4,6.
So, the number of favourable outcomes to the event getting an even number $=3$
Therefore, P(getting an even number) $=\frac{\text { Favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{6}=\frac{1}{2}$
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Question 73 Marks
The heights of 60 students of a class are given in the following table :
Height(in cm)150-155 155-158158-160160-165165-172
No. of students8 202543
Find the probability that the height of a student lies in the interval (iii) 158-172
Answer
The total number of students $=60$
Total number of students whose height lie in the interval $158-172=25+4+3=32$
P(Height of a student lies in the interval 158-172)
$=\frac{\text { Number of students whose height lies in the interval 158-172 }}{\text { Total number students }}$
$
=\frac{32}{60}=\frac{8}{15}
$
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Question 83 Marks
The heights of 60 students of a class are given in the following table :
Height(in cm)150-155155-158158-160160-165165-172
No. of students8202543
Find the probability that the height of a student lies in the interval (ii) 150-160
Answer
The total number of students $=60$
Total number of students whose height lie in the interval
$
150-160=8+20+25=53
$
P(Height of a student lies in the interval 150-160)
$
=\frac{\text { Number ot students whose height lies in the interval } 150-160}{\text { Total number students }}=\frac{53}{60}
$
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Question 93 Marks
The heights of 60 students of a class are given in the following table :
Height(in cm)150-155155-158158-160160-165165-172
No. of students8202543
Find the probability that the height of a student lies in the interval
(i) 155-158
Answer
The total number of students $=60$
Total number of students whose height lie in the interval $155-158=20$
P(Height of a student lies in the interval 155-158)
$
=\frac{\text { Number of students whose height lies in the interval 155-158 }}{\text { Total number students }}=\frac{20}{60}=\frac{1}{3}
$
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Question 103 Marks
The percentage of marks obtained by a student in monthly unit tests are given below:
Unit TestlllllllVVVl
Percentage of marks obtained726769747176
Based on this data find the probability that the student gets
(iii) less than 65% marks in a unit test.
Answer
The total number of tests taken by a student $=6$
The number of times a student gets less than $65 \%$ marks in a unit test $=0$
P(less than $65 \%$ marks in a unit test)
$=\frac{\text { Number of times student gets less than } 65 \% \text { marks in a unit test }}{\text { Total number of tests taken by a student }}$
$=\frac{0}{6}=0$
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Question 113 Marks
The percentage of marks obtained by a student in monthly unit tests are given below :
Unit TestlllllllVVVl
Percentage of marks obtained726769747176
Based on this data find the probability that the student gets
(ii) less than 72% marks in a unit test.
Answer
The total number of tests taken by a student $=6$
The Number of times a student gets less than $72 \%$ marks in a unit test $=3$
$
\begin{aligned}
& P(\text { less than } 72 \% \text { marks in a unit test) } \\
& =\frac{\text { Number of times student gets less than } 72 \% \text { marks in a unit test }}{\text { Total number of tests taken by a student }} \\
& =\frac{3}{6}=\frac{1}{2}
\end{aligned}
$
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Question 123 Marks
The percentage of marks obtained by a student in monthly unit tests are given below :
Unit TestlllllllVVVl
Percentage of marks obtained726769747176
Based on this data find the probability that the student gets
(i) More than 70% marks in a unit test.
Answer
The total number of tests taken by a student $=6$
The number of times a student gets more than $70 \%$ marks in a unit test $=4$
$
\begin{aligned}
& P(\text { More than } 70 \% \text { marks in a unit test) } \\
& =\frac{\text { Number of times student gets more than } 70 \% \text { marks in a unit test }}{\text { Total number of tests taken by a student }} \\
& =\frac{4}{6}=\frac{2}{3}
\end{aligned}
$
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Question 133 Marks
A die is thrown 450 times and frequencies of the outcomes 1,2,3,4,5,6 were noted as given in the following table
outcomes123456
Frequency737074758078
(iii) a number > 4
Answer
Total number of times die is thrown $=450$
Number of times greater than 4 come up $=80+78=158$
$
\mathrm{P}(>4 \text { will come up on die })=\frac{\text { Number of times }(>4) \text { come up }}{\text { Total number of times the die is thrown }}
$
$
=\frac{158}{450}=\frac{79}{225}
$
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Question 143 Marks
A die is thrown 450 times and frequencies of the outcomes 1,2,3,4,5,6 were noted as given in the following table
outcomes123456
Frequency737074758078
(ii) a number < 4
Answer
Total number of times die is thrown $=450$
Number of times less than 4 come up $=73+70+74=217$
$
\mathrm{P}(<4 \text { will come up on die })=\frac{\text { Number of times }(<4) \text { come up }}{\text { Total number of times the die is thrown }}
$
$
=\frac{217}{450}
$
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Question 153 Marks
A die is thrown 450 times and frequencies of the outcomes 1,2,3,4,5,6 were noted as given in the following table
outcomes123456
Frequency737074758078
(i) 4
Answer
Total number of times die is thrown $=450$
Number of times 4 come up on the die $=75$
$
\mathrm{P}(4 \text { will come up on die })=\frac{\text { Number of time } 4 \text { come up }}{\text { Total number of times the die is thrown }}
$
$
=\frac{75}{450}=\frac{1}{6}
$
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Question 163 Marks
1000 families with 2 children were selected randomly , and the following data were recorded :
Number of girls in a family012
Number of families333392275
Find the probability of a family , having (iii) no girl
Answer
Total number of families $=333+392+275=1000$
(iii) Number of families having no girl $=333$
$
\text { Required probability }=\frac{\text { Number of families having no girls }}{\text { Total number of families }}
$
$
=\frac{333}{1000}
$
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Question 173 Marks
1000 families with 2 children were selected randomly , and the following data were recorded :
Number of girls in a family012
Number of families333392275
Find the probability of a family , having
(ii) 2 girls
Answer
Total number of families $=333+392+275=1000$
(ii) Number of families having 2 girl $=275$
Required probability $=\frac{\text { Number of families having } 1 \text { girls }}{\text { Total number of families }}$
$
=\frac{275}{1000}=\frac{11}{40}
$
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Question 183 Marks
 1000 families with 2 children were selected randomly , and the following data were recorded :
Number of girls in a family012
Number of families333 392275
Find the probability of a family , having
(i) 1 girl 
Answer
Total number of families $=333+392+275=1000$
(i) Number of families having $1 \mathrm{girl}=392$
$
\begin{aligned}
& \text { Required probability }=\frac{\text { Number of families having } 2 \text { girls }}{\text { Total number of families }} \\
& =\frac{392}{1000}=\frac{49}{125}
\end{aligned}
$
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Question 193 Marks
Two dice are rolled simultaneously, write down the total number of possible outcomes.
Answer
When two dice are rolled the outcomes are :
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2. 1), (2. 2), (2. 3), (2. 4 ), (2. 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4 ), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4 ), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
So, the total number of outcomes = 36 
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Question 203 Marks
Two different dice are thrown at the same time. Find the probability of getting :
(iii) a sum of at least 11 
Answer
Total number of outcomes when two dice are rolled $=36$
A Sum at least 11
favorable outcomes are: $(5,6),(6,5),(6,6)$
Number of favorable outcomes $=3$
$
\mathrm{P}(\text { Sum is at least } 11)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{36}=\frac{1}{12}
$
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Question 213 Marks
Two different dice are thrown at the same time. Find the probability of getting :
(ii)sum divisible by 5
Answer
Total number of outcomes when two dice are rolled $=36$
Sum divisible by 5
favorable outcomes are: $(1,4),(2.3),(3,2),(4,1),(4,6),(5,5),(6,4)$
Number of favorable outcomes $=7$
$
P(\text { Sum divisible by } 5)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{7}{36}
$
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Question 223 Marks
Two different dice are thrown at the same time. Find the probability of getting :
(i) a doublet
Answer
Total number of outcomes when two dice are rolled $=36$
A doublet
favorable outcomes are: $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$
Number of favorable outcomes $=6$
$
P(\text { A doublet })=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{6}{36}=\frac{1}{6}
$
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Question 233 Marks
A box contains 5 red marbles, 8 white marbles and 4 grem marbles. One marble is taken out of the box at randam.
What is the probability that the marble taken out wil be
(iii) not green?
Answer
Total number of marbles $=5+8+4=17$
Number of green marbles $=4$
$
\text { Probability of getting a green marble }=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{4}{17}
$
Probability of not getting a green marble $=1-\frac{4}{17}=\frac{13}{17}$
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Question 243 Marks
There coins are tossed simultaneously . what is the probability of getting atleast two heads?
Answer
If three coins are tossed then possible outcomes are:
HHH , HHT, HTH , THH , HTT , THT , TTH , TTT
So total number of outcomes $=8$
Favorable out comes for getting at least 2 heads are :
HHH , HHT , HTH , THH
So number of favorable outcomes $=4$
Thus, $\mathrm{P}($ Getting at least 2 heads $)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{4}{8}=\frac{1}{2}$
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Question 253 Marks
A coin is tossed 800 times and the outcomes were noted as: Head:415 ,Tail:385
Find the probabiity of the min showing up (i)a head ,(ii) a tai .
Answer
As the coin is tossed 800 times, the total number of trials is 800 . Let $\mathrm{E}_1$ and $\mathrm{E}_2$ be the events of the coin coming up with a head and a tail respectively .
The number of times $E_1$ occurs $=415$ and the number of times $E_2$ occurs $=385$
$\therefore \mathrm{P}\left(\mathrm{E}_1\right)=\frac{415}{800}=\frac{83}{160}$, and
$
\mathrm{P}\left(\mathrm{E}_2\right)=\frac{385}{800}=\frac{77}{160}
$
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[3 marks sum] - Mathematics STD 10 Questions - Vidyadip