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Maths Case study (4 Marks)

Probability · CBSE STD 10 (CBSE - English Medium). 10 questions.

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Case study (4 Marks)

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Question 14 Marks
In a survey on holidays, 120 people were asked to state which type of transport they used on their last holiday. The following pie chart shows the results of the survey.
Image
(i) If one person is selected at random, find the probability that he/she travelled by bus or ship.
(ii) Which is most favourite mode of transport and how many people used it?
(iii) (a) A person is selected at random. If the probability that he did not use train is $\frac{4}{5}$ find the number of people who used train.
OR
(b) The probability that randomly selected person used aeroplans is $\frac{7}{60}$. Find the revenue collected by air company at the rate of ₹ 5,000 per person.
Answer
(i) Probability that a person travelled by bus or ship.
$\begin{array}{l}=\frac{\text { A rea of sector of sector angle } 36^{\circ}+\text { Area of sector of sector angle } 33^{\circ}}{\text { Area of circle }} \\ =\frac{\frac{36}{360} \times \pi r^2+\frac{33}{360} \times \pi r^2}{\pi r^2}=\frac{69}{360}=\frac{23}{120}\end{array}$
(ii) Most favourate mode of transport is car.
Probability that a person travelled by car $=\frac{\frac{177}{360} \times \pi r^2}{\pi r^2}=\frac{59}{120}$
∴ Number of persons who travelled by car $=\frac{59}{120} \times 120=59$
(iii) (a) Probability that a person travelled by train $=1-\frac{4}{5}=\frac{1}{5}$
∴ Number of people who travelled by train $=\frac{1}{5} \times 120=24$
OR
(b) Probability that a person travelled by aeroplane $=\frac{7}{60}$
∴ Number of persons who travelled by aeroplane $=\frac{7}{60} \times 120=14$
∴ Revenue collected by air company $=₹(14 \times 5000)=₹ 70,000$.
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Question 24 Marks
Answer
(i) The possible outcomes of the game are: RR, RB, RG, GR, GB, GG, YR, YB, YG
(ii) Total number of outcomes = 9. Purple can be made if the outcome is RB.
∴ Probability of making purple $=\frac{1}{9}$
(iii) We find that:
Probability of winning $₹ 10=\frac{1}{9}$ and, Probability of losing $₹ 5=\frac{8}{9}$
∴ Amount collected by the school from one participant $=₹\left(-10 \times \frac{1}{9}+5 \times \frac{8}{9}\right)=₹ \frac{10}{3}$
Hence, Amount collected by the school from 99 participants $=₹\left(\frac{10}{3} \times 99\right)=₹ 330$
(iv) Probability of winning $₹ 5=\frac{1}{9}$ and, Probability of loosing $₹ 5=\frac{8}{9}$
∴ Amount collected by the school from one participant $=₹ \left(-5 \times \frac{1}{9}+5 \times \frac{8}{9}\right)=₹ \frac{35}{9}$
Hence, Amount collected by the school from 99 participants $=₹\left(99 \times \frac{35}{9}\right)=₹ 385$
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Question 34 Marks
Computer-based learning (CBL) refers to any teaching methodology that makes use of computers for information transmission. At an elementary school level, computer applications can be used to display multimedia lesson. A survey was done on 1000 elementary and secondary schools of Assam and they were classified by the number of computers they had.
Image
Number of Computers1-1011-2021-5051-100101 and more
Number of Schools25020029018080

One school is chosen at random. Then:
(i) Find the probability that the school chosen at random has more than 100 computers.
(ii) Find the probability that the school chosen at random has 50 or fewer computers.
(iii) Find the probability that the school chosen at random has not more than 20 computers.
(iv) Find the probability that the school chosen at random has 10 or less than 10 computers.
Answer
(i) Total number of schools = 1000
Number of schools having more than 100 computers = 80
∴ Probability that the school chosen at random has more than 100 computers $=\frac{80}{1000}=0.08$
(i) Number of schools having 50 or fewer computers = 250 + 200 + 290 = 740
∴ Required probability $=\frac{740}{1000}=0.74$
(ii) Number of schools having not more than 20 computers = 250 + 200 = 450
∴ Required probability $=\frac{450}{1000}=0.45$
(iv) Number of schools having 10 or less than 10 computers = 250
∴ Required probability $=\frac{250}{1000}=0.25$

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Question 44 Marks
Answer
Out of 15 numbered pool balls, one ball can be drawn in 15 ways.
Total number of elementary events = 15
(i). There is exactly one ball bearing number 8. Therefore, a ball bearing number 8 can be drawn in one way.
∴ Favourable number of elementary events = 1
Hence, Probability that the drawn ball bears number $8=\frac{1}{15}$
(ii) There are 7 even numbered balls out of which one ball can be drawn in 7 ways
∴ Favourable number of elementary events = 7
Hence, probability that the drawn ball bears an even number $=\frac{7}{15}$
(iii) There are 5 balls bearing numbers which are multiple of 3
∴ Favourable number of elementary events = 5
Hence, probability that the drawn ball bears a number, which is multiple of 5 = $\frac{5}{15}=\frac{1}{3}$
(iv) There are 4 solid coloured balls bearing an even number 2, 4, 6, 8
∴ Favourable number of elementary events = 4
Hence, probability that the drawn ball is a solid coloured ball bearing an even number = $\frac{4}{15}$.
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Question 54 Marks
Figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 20 cm and each of the other bands is 10 cm wide. If a bowman shoots an arrow. Find the following probabilities.
Image
(i) The probability that the arrow hits in Gold scoring area, is
(a) $\frac{1}{25}$$\quad$ (b) $\frac{3}{25}$$\quad$ (c)$\frac{1}{5}$$\quad$ (d) $\frac{7}{25}$
(ii) The probability that it hits in Red scoring area, is
(a) $\frac{1}{25}$$\quad$ (b) $\frac{3}{25}$$\quad$ (c) $\frac{1}{5}$$\quad$ (d) $\frac{7}{25}$
(iii) The probability that it hits in Blue scoring area, is
(a) $\frac{9}{25}$$\quad$ (b) $\frac{7}{25}$$\quad$ (c) $\frac{3}{25}$ $\quad$ (d)$\frac{1}{5}$
(iv) The probability that it hits in Black scoring area, is
(a) $\frac{9}{25}$$\quad$ (b) $\frac{3}{25}$$\quad$ (c) $\frac{7}{25}$$\quad$ (d) $\frac{1}{25}$
Answer
(i)(a): Radius of the archery target = 50cm .
∴ Area of archery target $=\pi \times 50^2 cm^2=2500 \pi cm^2$
∴ Area of gold scoring region $=\pi \times 10^2 cm^2=100 \pi cm^2$
∴,Probability that the arrow hits in gold scoring region $=\frac{100 \pi}{2500 \pi}=\frac{1}{25}$
(ii) (b): Area of red scoring region $=\pi\left(20^2-10^2\right) cm ^2=300 \pi cm^2$
∴ Probability that the arrow hits in red scoring area $=\frac{300 \pi}{2500 \pi}=\frac{3}{25}$
(iii) (d): Area of blue scoring region $=\pi\left(30^2-20^2\right) cm ^2=500 \pi cm^2$
∴ Probability that the arrow hits in blue scoring region $=\frac{500 \pi}{2500 \pi}=\frac{1}{5}$
(iv) (c): Area of black scoring region $=\pi\left(40^2-30^2\right) cm ^2=700 \pi cm^2$
∴ Probability that the arrow hits in black scoring region $=\frac{700 \pi}{2500 \pi}=\frac{7}{25}$
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Question 64 Marks
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being selected and the age X of the selected student is recorded. Using the above information answer each of the following questions:
(i) If an Indian citizen becomes eligible to cast his vote after attaining 18 years of age, what is the probability that the selected student is an eligible voter?
(a) $\frac{2}{15}$$\quad$ (b) $\frac{2}{5}$$\quad$ (c) $\frac{1}{3}$$\quad$ (d) $\frac{7}{15}$
(ii) Probability of selecting a student who will be an eligible voter after one year, is
(a) $\frac{2}{5}$$\quad$ (b) $\frac{3}{5}$$\quad$ (c) $\frac{1}{5}$$\quad$ (d) $\frac{1}{3}$
(iii) Probability of selecting a non-eligible voter is
(a) $\frac{7}{15}$$\quad$ (b) $\frac{8}{15}$$\quad$ (c) $\frac{3}{5}$$\quad$ (d) $\frac{1}{5}$
(iv) $P(X \geq 20)$ =
(a) $\frac{1}{5}$$\quad$ (b) $\frac{7}{15}$$\quad$ (c) $\frac{4}{15}$$\quad$ (d) $\frac{1}{5}$
Answer
(i) (d). Total number of elementary events = 15
There are 7 students of age 18 or more years
Favourable number of ways = 7
Hence, required probability = $\frac{7}{15}$
(ii) (c): A student will be eligible to his vote after one year, if he is 17 years old now There are three students who are 17 years old now.
Favourable number of elementary events = 3
Probability of selecting an eligible voter = $\frac{3}{15}$ = $\frac{1}{5}$
(iii) (b): Required probability = 1 - P (selecting an eligible voter) $=1-\frac{1}{5}=\frac{4}{5}$
(iv) (c): Gamma(X >= 20) = Probability of selecting a student of age 20 years or more = $\frac{4}{15}$
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Question 74 Marks
There is a game of tossing of three coins. The entry fee in the game is 200. Shikha takes entry into the game by paying the entry fee. The rules of the game are: If by tossing three coins or one coin three times she gets one or two heads entry fee is refunded. If she gets three heads, she receives double the amount paid as entry fee. Otherwise, she loses the entry fee.
Image
(i) The total number of possible outcomes of the game, is
(a) 4$\quad$ (b) 2$\quad$ (c) 6$\quad$ (d) 8
(ii) The probability that Shikha loses the entry fee, is
(a) $\frac{1}{8}$$\quad$ (b) $\frac{3}{8}$$\quad$ (c) $\frac{3}{4}$$\quad$ (d) $\frac{1}{2}$
(iii) What are the chances of getting ₹400?
(a) 50%$\quad$ (b) 25%$\quad$ (c) 12.5%$\quad$ (d) 75%
(iv) The probability that she gets back her entry fee, is
(a) $\frac{1}{8}$$\quad$ (b) $\frac{3}{4}$$\quad$ (c) $\frac{1}{8}$ $\quad$ (d) $\frac{3}{8}$
Answer
(i) (d): Following are 8 possible outcomes in tossing three coins. HHH, HHT, HTH. THH. TTH, HTT, THT, TTT.
(ii) (a): Shikha looses entry fee, if she gets all tails.
So, required probability = $\frac{1}{8}$
(ii) (c): Shikha will get double the amount paid as entry fee i.e. ₹400, if she gets all heads.
So, required probability = $\frac{1}{8}$
Hence, the chances are $\frac{1}{8} \times 100 \%=12.5 \%$
(iv) (b): Shikha gets back her gets one of the following as outcome:
HHT, HTH, THH, HTT, THT, TTH.
So, required probability = $\frac{6}{8}$ = $\frac{3}{4}$
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Question 84 Marks
Mira has a circular cardboard with centre O. The cardboard has been divided into three regions labelled X, Y and Z by diameter AB and radius OC as shown in the following figure. She places a fair spinner at the centre O of the cardboard. If angle BOC = $45^{\circ}$; answer each of the following questions:
Image
(i) The probability that the spinner will land in the region X, is
(a) $\frac{1}{4}$ $\quad$ (b) $\frac{1}{2}$$\quad$ (c) $\frac{1}{8}$$\quad$ (d) $\frac{3}{8}$
(ii) The probability that the spinner will land in the region Y, is
(a) $\frac{1}{4}$ $\quad$ (b) $\frac{1}{2}$$\quad$ (c) $\frac{1}{8}$$\quad$ (d) $\frac{3}{8}$
(iii) The probability that the spinner will land in the region Z, is
(a) $\frac{1}{4}$ $\quad$ (b) $\frac{1}{2}$$\quad$ (c) $\frac{1}{8}$$\quad$ (d) $\frac{3}{8}$
(iv) The probability that the spinner will land in the region X or Z, is
(a) $\frac{1}{4}$ $\quad$ (b) $\frac{1}{2}$$\quad$ (c) $\frac{1}{8}$$\quad$ (d) $\frac{3}{8}$
Answer
(i) (d). Letr be the radius of the circular card-board. Then,
Area of the card-board $=\pi r^2$
Area of region $X=\frac{135}{360} \times \pi r^2=\frac{3}{8} \pi r^2$
∴ Required probability$=\frac{\frac{3}{8} \pi r^2}{\pi r^2}=\frac{3}{8}$
(ii)  (b): Area of region $\gamma=\frac{\pi}{2} r^2$
∴ Required probability $Y=\frac{\frac{\pi}{2} r^2}{\pi r^2}=\frac{1}{2}$
(iii) (c): Area of region $Z=\frac{45}{360} \pi r^2=\frac{\pi}{8} r^2$
∴ Required probability $=\frac{8}{\pi r^2}=\frac{1}{8}$
(iv) (b): Total area of region X and Z $=\frac{\pi r^2}{2}$
∴ Required probability $=\frac{\frac{\pi r^2}{2}}{\pi r^2}=\frac{1}{2}$
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Question 94 Marks
A missing helicopter is reported to have crashed somewhere in the rectangular region in Fig. 15.2. In the rectangular region there is a lake in one corner and in the remaining corners there are three water bodies each in the shape of a quadrant of a circle of radius 700 m.
Image
(i) The probability that the helicopter crashed inside the lake is
(a) 0.1$\quad$ (b) 0.4$\quad$ c) 0.5$\quad$ (d) 0.8
(ii) What is the chance of helicopter crashing in one of the three water bodies in three corners?
(a) 2%$\quad$ (b) 2.88%$\quad$ c) 4%$\quad$ (d)3.88%
(iii) The probability that the helicopter crashed in the water body near to the lake, is
(a) $\frac{77}{4000}$$\quad$ (b) $\frac{77}{2000}$$\quad$ (c) $\frac{77}{8000}$$\quad$ (d) $\frac{231}{8000}$
(iv) The probability that the helicopter crashed in a water body including lake is
(a) $\frac{231}{8000}$$\quad$ (b) $\frac{1031}{8000}$$\quad$ c)$\frac{931}{8000}$$\quad$ (d)$\frac{1037}{8000}$
Answer
(i) (a): Area of the entire region $=(8 \times 5) km ^2=40 km^2$
Area of the lake $=(2 \times 2) km ^2=4 km^2$
∴ Probability that the helicopter crashed inside the lake = $\frac{4}{40}=\frac{1}{10}$
(ii) (b): Each water body is a quadrant of a circle of radius $\frac{700}{1000}$= 0.7km
Area of 3 water bodies $=\frac{3}{4} \times \frac{22}{7} \times(0.7)^2 km^2=\frac{231}{200} km^2$
∴ Probability that the helicopter crashed in one of the water bodies $=\frac{231}{200 \times 40}=\frac{231}{8000}=0.0288$
Chances of helicopter crashing in one of the water bodies = 2.88%
(iii) (c): Area of the water body nearer to the lake $=\frac{1}{4}\left(\frac{22}{7} \times 0.7 \times 0.7\right)=\frac{154}{400} km^2$
∴ Probability that the helicopter crashes in the water body nearer to the lake $=\frac{154}{400 \times 40}=\frac{77}{8000}$
(iv) (c): Area of three water bodies and lake $=\left(\frac{3}{4} \times \frac{22}{7} \times(0.7)^2+2 \times 2\right) km ^2=\frac{1031}{200} km^2$
∴ Probability that the helicopter crashes in a water body or in lake $=\frac{1031}{200 \times 40}=\frac{1031}{8000}$
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Question 104 Marks
In a game of chance consisting of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, ..., 20 (see Fig. 15.1) and these are equally likely outcomes. Three persons Aarushi, Avni and Mira decide to play the game. Aarushi wins ₹ 5000 if the arrow points at an even number. Mira wins ₹ 8000 when arrow points at an odd number and Avni wins ₹ 12,000 when arrow points at a prime number.
Image
(i) Total number of possible outcomes in the game is:
(a) 10$\quad$ (b) 20$\quad$ (c) 30$\quad$ (d)40
(ii) What are the chances of Aarushi winning ₹5000?
(a) 50%$\quad$ (b) 60%$\quad$ (c) 40%$\quad$ (d) 25%
(iii) What are the chances of Mira winning *8000?
(a) 60%$\quad$ (b) 40%$\quad$ (c) 50%$\quad$ (d) 80%
(iv) What are the chances of Avni winning 12000?
(a) 50%$\quad$ (b) 60%$\quad$ (c) 80%$\quad$ (d) 40%
(v) What are the chances of Mira and Avni winning their prize money?
(a) 35%$\quad$ (b) 5%$\quad$ (c) 40%$\quad$ (d) 25%
Answer
(i) (b): The spinning arrow may come to rest at any one of 20 numbers. So, total number of possible outcomes is 20.
(ii) (a): There are 10 even numbers in numbers from 1 to 20. So, Aarushi can win if the spinning arrow stops at any one of the even numbers.
∴ Favourable number of outcomes = 10.
Probability that Aarushi wins $=\frac{10}{20}=\frac{1}{2} \Rightarrow$ Chances of Aarushi's win $=\frac{1}{2} \times 100 \%=50 \%$
(iii) (c): There are 10 odd numbers and Meera can win if the spinning arrow stops at any one of the odd numbers.
∴ Favourable number of outcomes = 10.
Probability that Meera wins $=\frac{10}{20}=\frac{1}{2} \Rightarrow$ Chances of Meera's win $=\frac{1}{2} \times 100 \%=50 \%$
(iv) (d): Avni wins if the arrow stops at a prime number. There are 8 prime numbers viz. 2, 3, 5, 7, 11, 13, 17, 19 between 1 to 20.
∴ Favourable number of outcomes = 8.
Probability that Avni wins $=\frac{8}{20}=\frac{1}{2} \Rightarrow$ Chances of Avni's win $=\frac{2}{5} \times 100 \%=40 \%$
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Case study (4 Marks) - Maths STD 10 Questions - Vidyadip