(i)\(7\over24\)
(ii)\(5\over6\)

Total number of tickets 1000, Number tickets with prizes = 5
\(\therefore\) Probability of winning a prize = \(\frac{5}{1000}\)
  = 0.005

Total number of cards = 52 and number of Jack of hearts  =1
\(\therefore\)  Probability of drawing a Jack of hearts = \(\frac{1}{52}\)

Let number of bad pens = x.
\(\therefore\)  Probability of getting a bad pen = \(\frac{x}{400}\)
A.T.Q. \(\frac{x}{400}\) =0.25
\(\Rightarrow\)   x = 0.25 x 400 = 100
\(\therefore\)  Number of good pens = 400 - 100 = 300

Leap year contains 366 days.  \(\Rightarrow \)  52 weeks + 2 days
52 weeks contain 52 Fridays
We will get 53 Fridays if one of the remaining two days is a Friday. Total possibilities for two days are:
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
There are 7 possibilities and out of these there are 2 favourable cases.
\(\therefore P(53 Fridays) = \frac{2}{7}\)

Total no.of cards = 46
Total no.of ways to select a card = 46
(i) Prime no.less than 10 in these cards are 5, 7
\(\therefore\)  No.of ways to select a prime no.less than 10 = 2
\(\therefore\)  Probability that the number on the card is prime = \(\frac{2}{46}=\frac{1}{23}\)
(ii) No. which is a perfect square, i.e. 9, 16, 25, 36, 49
No. of ways to select a card with perfect square = 5
\(\therefore\)   Probability = \(\frac{5}{46}\)

Area of restangle = 3 \(\times\) 2 = 6 m²
and area of circle of radius \(\frac{1}{2} \mathrm{~m}=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4} \mathrm{~m}^2\)
\(\left[\because \text { diameter }=1 \mathrm{~m} \Rightarrow \text { radius }=\frac{1}{2} \mathrm{~m}\right]\)
Now, probanility that the die land inside the circle
\(=\frac{\text { Area of circle }}{\text { Area of rectangle }}=\frac{\pi / 4}{6}=\frac{\pi}{24}\)

(i) The car starts normally but when there is some defect, then car does not start. So, the outcomes are not equally likely.
(ii) The outcomes in this situation are not equally likely because the outcomes depends on many factors such as training ofthe player, quality of basketball, etc.
(iii) The outcomes in trial of true-false question is either true or false. Hence, the two outcomes are equally likely.
(iv) A new baby can be either a boy or a girl, so both the outcomes are equally likely.

Total number of fishes in the tank = 5 male fishes + 8 female fishes = 13 fishes
\(\therefore\) Probability of taken out a male fish
\(=\frac{\text { Number of male fishes }}{\text { Total number of fishes }}=\frac{5}{13}\)

Given, number of 50 paise coins = 100,
number of Rs 1 coins = 50,
number of Rs 2 coins = 20
and number of Rs 5 coins = 10
\(\therefore\) total number of coins = 100 + 50 + 20 + 10 = 180

(i) P (50 paise coin)=\(\begin{aligned} & \frac{\text { Number of } 50 \text { paise coins }}{\text { Total number of coins }} \\ \end{aligned}\)
=\(\begin{aligned} \frac{100}{180}=\frac{5}{9} \end{aligned}\)
(ii) Number of coins which are not of Rs 5
= Total number of coins - Number of Rs 5 coins = 180 - 10 = 170
\(\therefore\) P (that the coin will not be a Rs 5 coin) \(\begin{aligned} & =\frac{170}{180} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{17}{18} \end{aligned}\)

Number of red balls = 3
Number of black balls = 5
Total number of balls = 3 + 5 = 8
(i) P(red ball) = \(\frac{Number \ \ of \ \ red \ \ balls}{Total \ \ number \ \ of \ \ balls}=\frac{3}{8}\)
(ii) P(not red) = \(1-\frac{3}{8}=\frac{5}{8}\)

The helicopter is equally likely to crash anywhere in the region.
Area of the entire region where the helicopter can crash
= (4.5 x 9) km² = 40.5 km²
Area of the lake = (2.5 x 3) km² = 7.5 km²
Therefore, P (helicopter crashed in the lake) = \(\frac{7.5}{40.5}=\frac{75}{405}=\frac{5}{27}\)

Saying that a marble is drawn at random is a short way of saying that all the marbles are equally likely to be drawn. Therefore, the
number of possible outcomes = 3 +2 + 4 = 9 (Why?)
Let W denote the event ‘the marble is white’, B denote the event ‘the marble is blue' and R denote the event ‘marble is red’.
(i) The number of outcomes favourable to the event W = 2
So, P(W) \(=\frac{2}{9}\)
Similarly, (ii) P(B)  \(=\frac{3}{9}=\frac{1}{3} \ \text { and } \ (\text { iii }) P(R)=\frac{4}{9}\)
Note that P(W) + P(B) + P(R) = 1

(i) Total possible outcomes on throwing two dice are
(1,1), (1, 2), (1,3), (1, 4), (1,5), (1,6)
(2,1),(2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1),(4, 2), (4, 3),(4, 4),(4, 5), (4, 6)
(5, 1),(5, 2), (5, 3), (5, 4), (5, 5),(5, 6)
(6,1), (6, 2), (6, 3), (6, 4), (6, 5) and (6, 6)
\(\therefore\) Number of all possible outcomes = 36
a) Let E₁ = Sum of two dice is 3.
Then, E₁ would consist of two outcomes, namely (1, 2) and (2, 1).
\(\therefore\) Number of outcomes favourable to E₁ = 2
Hence, \(P\left(E_1\right)=\frac{2}{36}=\frac{1}{18}\)
(b) Let E₂ = Sum of two dice is 4.
Then, E₂ would consist of three outcomes, namely (1, 3), (2, 2) and (3, 1).
\(\therefore\) Number of outcomes favourable to E₂ = 3
Hence, \(P\left(E_2\right)=\frac{3}{36}=\frac{1}{12}\)
(c) Let E₃ = Sum of two dice is 5.
Then, E₃ would consist of four outcomes, namely (1, 4), (2, 3), (3, 2) and (4, 1).
\(\therefore\) Number of outcomes favourable to E₃ = 4
Hence, P(E₃) = \(\frac{4}{36}=\frac{1}{9}\)
(d) Let E₄ = Sum of two dice is 6.
Then, E₄ would consist of five outcomes, namely (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1).
\(\therefore\) Number of outcomes favourable to E₄ = 5
Hence, P(E₄) = \(\frac{5}{36}\)
(e) Let E₅ = Sum of two dice is 7.
then, E₅ would consist of six outcomes, namely (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1).
\(\therefore\) Number of outcomes favourable to E₅ = 6
Hence, P(E₅) = \(\frac{6}{36}=\frac{1}{6}\)
(f) Let E₆ = Sum of two dice is 9.
Then, E₆ would consist of four outcomes, namely (3, 6), (4, 5), (5, 4) and (6, 3).
 \(\therefore\) Number of outcomes favourable to E₆ = 4
Hence, P(E₆) = \(\frac{4}{36}=\frac{1}{9}\)
 (g) Let E₇ = Sum of two dice is 10.
Then, E₇ would consist of three outcomes, namely (4, 6), (5, 5) and (6, 4).
\(\therefore\) Number of outcomes favourable to E₇ = 3
Hence, P(E₇) = \(\frac{3}{36}=\frac{1}{12}\)
(h) Let E₈ = Sum of two dice is 11.
Then, E₈ would consist of two outcomes, namely (6, 5) and (5, 6).
\(\therefore\) Number of outcomes favourable to E₈ = 2
Hence, P(E₈) = \(\frac{2}{36}=\frac{1}{18}\)
(ii) No, we do not agree with the given argument because the events of eleven sumd are not equally likely.

Total number of cards in one deck of cards is 52.
\(\therefore\) Total number of outcomes =52
(i) Let E₁ = Event of getting a king of red colour
\(\therefore\) Number of outcomes favourable to E₁ =2
[\(\because\) there are four kings in a deck of playing cards out of which two are red and two are black]
Hence, probability of getting a king of red colour,
\(P\left(E_1\right)=\frac{2}{52}=\frac{1}{26}\)
(ii) Let E₂ = Event of getting a face card
\(\therefore\) Number of outcomes favourable to E₂ =12
(\(\because\) in a deck of cards, there are 12 face cards, namely 4 kings, 4 jacks, 4 queens]
Hence, probability of getting a face card,
\(P\left(E_2\right)=\frac{12}{52}=\frac{3}{13}\)
(iii) P(a red face card)  \(\frac{6}{52}\)  = \(\frac{3}{26}\)
(iv) Let E₄= Event of getting a jack of heart
\(\therefore\)Number of outcomes favourable to E₄ = 1 
[\(\because\) there are four jack cards in a deck, namely 1 of heart, 1 of club, 1 of spade and 1 of diamond] 
Hence, probability of getting a jack of heart,
\(P\left(E_4\right)=\frac{1}{52}\)
(v) Let E₅ = Event of getting a spade
\(\therefore\) Number of outcomes favourable to E₅ = 13
[\(\because\) in a deck of cards, there are 13 spades, 13 clubs, 13 hearts and 13 diamonds]
Hence, probability of getting a spade,
\(P\left(E_5\right)=\frac{13}{52}=\frac{1}{4}\)
(vi) P(the queen of diamonds) =\(\frac{1}{52}\)