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:

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})