Probability — Case-Study Based Question

A square dartboard has sections numbered from 1 to 12 as shown below. Players have to make a prediction and throw a dart. They win if their dart lands on the section that matches their prediction. Three friends Amit, Aryan and Aman took the challenge.

Based on the above information, answer the following :

(i) What is the probability of Amit winning if he predicted that his dart will land on a prime number ?

(ii) What is the probability of Aryan winning if he predicted that his dart will land on an odd number ?

(iii) What is the probability of Aman winning if he predicted that his dart will land on a square number ?

(iv) Which of the three friends has the highest chance of winning ?

Answer

The dartboard has 12 equally likely sections numbered 1 to 12, so the total number of outcomes = 12.

By formula,

Probability of an event = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

(i) Prime numbers from 1 to 12 are 2, 3, 5, 7 and 11, i.e. 5 numbers.

∴ P(Amit wins) = $\dfrac{5}{12}$.

Hence, the probability of Amit winning = $\dfrac{5}{12}$.

(ii) Odd numbers from 1 to 12 are 1, 3, 5, 7, 9 and 11, i.e. 6 numbers.

∴ P(Aryan wins) = $\dfrac{6}{12} = \dfrac{1}{2}$.

Hence, the probability of Aryan winning = $\dfrac{1}{2}$.

(iii) Square numbers from 1 to 12 are 1, 4 and 9, i.e. 3 numbers.

∴ P(Aman wins) = $\dfrac{3}{12} = \dfrac{1}{4}$.

Hence, the probability of Aman winning = $\dfrac{1}{4}$.

(iv) Comparing the three probabilities :

$\dfrac{5}{12} \approx 0.42, \quad \dfrac{1}{2} = 0.5, \quad \dfrac{1}{4} = 0.25$

Since $\dfrac{1}{2}$ is the largest, Aryan has the highest chance of winning.

Hence, Aryan has the highest chance of winning.

A teacher conducted a survey for 180 students of class 10 for a mathematics project to be given in the class.

She asked the students two MCQ questions, which were "What is their favourite subject ?" and "What time do they use their mobile phone at night ?". Each student could choose only one option from the choices given below.

The following are the results of the survey conducted :

Based on the above information, answer the following questions :

(i) If a student is selected at random, what is the probability that the student uses mobile phone after 11 pm and that their favourite subject is Mathematics ?

(ii) If a student is selected at random, what is the probability that the student uses mobile phone after 10 pm ?

(iii) If a student is selected at random, what is the probability that their favourite subject is Science ?

(iv) If a student is selected at random, what is the probability that the student uses mobile phone before 10 pm and that their favourite subject is either Mathematics or Computer ?

Answer

Total number of students = 180.

By formula,

Probability of an event = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

(i) Number of students who use mobile phone after 11 pm and whose favourite subject is Mathematics = 22.

∴ Required probability = $\dfrac{22}{180} = \dfrac{11}{90}$.

Hence, the required probability = $\dfrac{11}{90}$.

(ii) Students who use mobile phone after 10 pm = (10 pm - 11 pm) + (After 11 pm)

= (15 + 18 + 6 + 10) + (22 + 10 + 6 + 8)

= 49 + 46 = 95.

∴ Required probability = $\dfrac{95}{180} = \dfrac{19}{36}$.

Hence, the required probability = $\dfrac{19}{36}$.

(iii) Number of students whose favourite subject is Science = 12 + 15 + 18 + 10 = 55.

∴ Required probability = $\dfrac{55}{180} = \dfrac{11}{36}$.

Hence, the required probability = $\dfrac{11}{36}$.

(iv) Students who use mobile phone before 10 pm are those in the rows 'Before 9 pm' and '9 pm - 10 pm'.

Number of such students whose favourite subject is Mathematics or Computer = (20 + 8) + (13 + 5)

= 28 + 18 = 46.

∴ Required probability = $\dfrac{46}{180} = \dfrac{23}{90}$.

Hence, the required probability = $\dfrac{23}{90}$.