GSEB Class 12 Statistics Solutions Chapter 1 Probability Exercise 1.5

 

Question 1. The sample data about monthly travel expense (in Rs.) of a large group of travellers of local bus in a megacity are given in the following table:

Monthly travel expense (Rs.)	501-600	601-700	701-800	801-900	901 or more
No. of travellers	318	432	639	579	174

One person from this megacity travelling by local bus is randomly selected. Find the probability that the monthly travel expense of this person will be
(1) more than Rs. 900
(2) at the most Rs. 700
(3) Rs. 601 or more but Rs. 900 or less.
Answer: To begin, we calculate the total number of travellers in the sample, which is \(n = 318 + 432 + 639 + 579 + 174 = 2142\).
(1) Let A be the event that a person's monthly travel expense is more than Rs. 900.
The number of travellers whose monthly travel expense is more than Rs. 900 is 174.
So, the probability \(P(A)\) is the relative frequency of these travellers.
\(P(A) = \frac{\text{No. of travellers whose monthly travel expense is more than Rs. 900}}{\text{Total no. of travellers}}\)
\(P(A) = \frac{m}{n} = \frac{174}{2142}\)
\(P(A) = \frac{29}{357}\)
(2) Let B be the event that a person's monthly travel expense is at most Rs. 700.
This includes expenses from 501-600 and 601-700.
The number of travellers whose monthly travel expense is at most Rs. 700 is \(318 + 432 = 750\).
So, the probability \(P(B)\) is the relative frequency of these travellers.
\(P(B) = \frac{\text{No. of travellers whose monthly travel expense is at the most Rs. 700}}{\text{Total no. of travellers}}\)
\(P(B) = \frac{m}{n} = \frac{318+432}{2142} = \frac{750}{2142}\)
\(P(B) = \frac{125}{357}\)
(3) Let C be the event that a person's monthly travel expense is Rs. 601 or more but Rs. 900 or less.
This includes expenses from 601-700, 701-800, and 801-900.
The number of travellers whose monthly travel expense is Rs. 601 or more but Rs. 900 or less is \(432 + 639 + 579 = 1650\).
So, the probability \(P(C)\) is the relative frequency of these travellers.
\(P(C) = \frac{\text{No. of travellers whose monthly travel expense is Rs. 601 or more but Rs. 900 or less}}{\text{Total no. of travellers}}\)
\(P(C) = \frac{m}{n} = \frac{432+639+579}{2142} = \frac{1650}{2142}\)
\(P(C) = \frac{275}{357}\)
In simple words: We calculated the total number of travellers first. Then, for each part, we counted how many travellers met the given expense condition and divided that count by the total to find the probability.

🎯 Exam Tip: Always clearly define the event for each probability calculation and ensure all relevant categories are included in the numerator. Simplify fractions to their lowest terms for full marks.

 

Question 2. The details of a sample inquiry of 4979 voters of constituency are as follows:
(1) One voter is randomly selected from this constituency. If the voter is male, find the probability that he is a supporter of party A.
(2) If this voter is a supporter of party A, find the probability that he is a male.
Answer: Here, the total number of voters surveyed is \(n = 4979\).
Let A be the event that the selected voter is male.
To find the number of male voters, we would typically sum the males from various categories. Assuming data from a table (not provided in OCR, but implied by calculation \(1319 + 1217\)):
Number of male voters \(m = 1319 + 1217 = 2536\).
The probability of selecting a male voter is:
\(P(A) = \frac{\text{No. of male voters}}{\text{Total no. of voters}}\)
\(P(A) = \frac{m}{n} = \frac{2536}{4979}\)
Let B be the event that the selected voter supports party A.
To find the number of voters supporting party A, we would sum them from various categories (implied by calculation \(1319 + 1118\)):
Number of voters supporting party A \(m = 1319 + 1118 = 2437\).
The probability of selecting a voter who supports party A is:
\(P(B) = \frac{\text{No. of voters who are supporters of party A}}{\text{Total no. of voters}}\)
\(P(B) = \frac{2437}{4979}\)
Let \(A \cap B\) be the event that the selected voter is both male and a supporter of party A.
From the implied data (likely a common cell from a table), the number of voters who are male and supporters of party A is 1319.
The probability of this combined event is:
\(P(A \cap B) = \frac{\text{No. of voters favourable for event A and B}}{\text{Total no. of voters}}\)
\(P(A \cap B) = \frac{m}{n} = \frac{1319}{4979}\)
(1) We need to find the probability that the voter is a supporter of party A, *given* that the voter is male (P(B|A)).
\(P(B|A) = \frac{P(A \cap B)}{P(A)}\)
\(P(B|A) = \frac{\frac{1319}{4979}}{\frac{2536}{4979}}\)
\(P(B|A) = \frac{1319}{2536}\)
(2) We need to find the probability that the voter is male, *given* that the voter is a supporter of party A (P(A|B)).
\(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
\(P(A|B) = \frac{\frac{1319}{4979}}{\frac{2437}{4979}}\)
\(P(A|B) = \frac{1319}{2437}\)
In simple words: We calculated the total voters, then found probabilities for a voter being male, supporting Party A, and being both. Using these, we determined conditional probabilities: first, the chance of supporting Party A if male, and second, the chance of being male if supporting Party A.

🎯 Exam Tip: Clearly define all events (A, B, A ∩ B) and state the formula for conditional probability \(P(X|Y) = \frac{P(X \cap Y)}{P(Y)}\). Ensure you correctly identify the 'given' event for the denominator.