Gujarat Board Statistics Class 12 GSEB Solutions Part 2 Chapter 1 Probability Ex 1.5 Textbook Exercise Questions and Answers.
Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1.5
Question 1.
The sample data about monthly travel expense (in?) of a large group of travellers of local bus in a megacity are given in the following table:
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 ₹ 900
(2) at the most ₹ 700
(3) ₹ 601 or more but ₹ 900 or less.
Answer:
Here, the number of travellers selected in the sample is n = 318 + 432 + 639 + 579 + 174 = 2142
(1) A = Event that the monthly travel expense of the persons is more than ₹ 900.
∴ P(A) = Relative frequency of the number of travellers whose monthly travel expense is more than ₹ 900
= \(\frac{\text { No. of travellers whose monthly travel expense is more than } ₹ 900}{\text { Total no. of travellers }}\)
= \(\frac{m}{n}\) = \(\frac{174}{2142}\) = \(\frac{29}{357}\)
(2) B = Event that the monthly travel expense of the persons is at the most ₹ 700.
∴ P(B) = Relative frequency of the number of travellers whose monthly travel expense is at the most ₹ 700
= \(\frac{\text { No. of travellers whose monthly travel expense is at the most } ₹ 700}{\text { Total no. of travellers }}\)
= \(\frac{m}{n}\) = \(\frac{318+432}{2142}\) = \(\frac{750}{2142}\) = \(\frac{125}{357}\)
(3) C = Event that the monthly travel expense of the persons is ₹ 601 or more but ₹ 900 or less.
∴ P (C) = Relative frequency of the number of travellers whose monthly travel expense is ₹ 601 or more but ₹ 900 or less
= \(\frac{\text { No. of travellers whose monthly travel expense is } ₹ 601 \text { or more but } ₹ 900 \text { or less }}{\text { Total no. of travellers }}\)
= \(\frac{m}{n}\) = \(\frac{432+639+579}{2142}\) = \(\frac{1650}{2142}\) = \(\frac{275}{357}\)
Question 2.
The details of a sample inquiry of 4979 voters of constituency are as follows:
| Year | Time t | ŷ = 177.8 + 5.6t |
| 2011 | 1 | 177.8 + 5.6 (1) = 183.4 |
| 2012 | 2 | 177.8 + 5.6(2) = 189.0 |
| 2013 | 3 | 177.8 + 5.6 (3) = 194.6 |
| 2014 | 4 | 177.8 + 5.6 (4) = 200.2 |
| 2015 | 5 | 177.8 + 5.6 (5) = 205.8 |
One voter is randomly selected from this constituency.
(1) If this voter is a 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 numbers of voters is n = 4979
A = Event that the selected voter is a male.
∴ P(A) = Relative frequency of male voters
= \(\frac{\text { No. of male voters }}{\text { Total no. of voters }}\)
= \(\frac{m}{n}\)
Putting, m = 1319 + 1217 = 2536 and n = 4979
P(A) = \(\frac{2536}{4979}\)
B = Event that the selected voter is supporter of party A.
∴ P(B) = Relative frequency of voters who are supporters of party A
= \(\frac{\text { No. of voters who are supporters of party A }}{\text { Total no. of voters }}\)
= \(\frac{2437}{4979}\)
Putting, m = 1319 + 1118 = 2437 and n = 4979
P(B) = \(\frac{2437}{4979}\)
A ∩ B = Event that the selected voter is a male and supporter of party A .
∴ P(A ∩ B) = Relative frequency of the event A ∩ B
= \(\begin{aligned}
&\text { No. of voters favourable for }\\
&=\frac{\text { the event } A \cap B}{\text { Total no. of voters }}
\end{aligned}\)
= \(\frac{m}{n}\)
= \(\frac{1319}{4979}\)
(1) B|A = Event that the voter is a male that he is a supporter of party A
∴ P(B|A) = \(\frac{P(A \cap B)}{P(A)}\)
= \(\frac{\frac{1319}{4979}}{\frac{2536}{4979}}\)
= \(\frac{1319}{2536}\)
(2) A|B = Event that the voter is a supporter of party A that he is a male.
∴ P(A|B) = \(\frac{P(A \cap B)}{P(A)}\)
= \(\frac{\frac{1319}{4979}}{\frac{2437}{4979}}\)
= \(\frac{1319}{2437}\)