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 â‚¹ 900

(2) at the most â‚¹ 700

(3) â‚¹ 601 or more but â‚¹ 900 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 â‚¹ 900.

âˆ´ P(A) = Relative frequency of the number of travellers whose monthly travel expense is more than â‚¹ 900

= \(\frac{\text { No. of travellers whose monthly travel expense is more than } â‚¹ 900}{\text { Total no. of travellers }}\)

= \(\frac{m}{n}\) = \(\frac{174}{2142}\) = \(\frac{29}{357}\)

(2) B = Event that the monthly travel expense of the persons is at the most â‚¹ 700.

âˆ´ P(B) = Relative frequency of the number of travellers whose monthly travel expense is at the most â‚¹ 700

= \(\frac{\text { No. of travellers whose monthly travel expense is at the most } â‚¹ 700}{\text { Total no. of travellers }}\)

= \(\frac{m}{n}\) = \(\frac{318+432}{2142}\) = \(\frac{750}{2142}\) = \(\frac{125}{357}\)

(3) C = Event that the monthly travel expense of the persons is â‚¹ 601 or more but â‚¹ 900 or less.

âˆ´ P (C) = Relative frequency of the number of travellers whose monthly travel expense is â‚¹ 601 or more but â‚¹ 900 or less

= \(\frac{\text { No. of travellers whose monthly travel expense is } â‚¹ 601 \text { or more but } â‚¹ 900 \text { or less }}{\text { Total no. of travellers }}\)

= \(\frac{m}{n}\) = \(\frac{432+639+579}{2142}\) = \(\frac{1650}{2142}\) = \(\frac{275}{357}\)

Question 2.

The details of a sample inquiry of 4979 voters of constituency are as follows:

Year | Time t | yÌ‚ = 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.

âˆ´ 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.

âˆ´ 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 âˆ© B = Event that the selected voter is a male and supporter of party A .

âˆ´ P(A âˆ© B) = Relative frequency of the event A âˆ© 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

âˆ´ 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.

âˆ´ P(A|B) = \(\frac{P(A \cap B)}{P(A)}\)

= \(\frac{\frac{1319}{4979}}{\frac{2437}{4979}}\)

= \(\frac{1319}{2437}\)