Total outcomes = 36
(i) let A = a prime number on each die.
Favourable outcomes of event A are (2, 2), (2,3), (2, 5), (3, 3), (3, 2), (3, 5),(5,2),(5,3),(5,5)
\(\therefore\)P(A)=\(\frac{9}{36}=\frac{1}{4}\)
(ii) let B = a total of 9 or 11
Favourable outcomes of event B are 
(3,6),(4,5),(5,4),(5,6),(6,3),(6,5)
\(\therefore\)P(B)=\(\frac{6}{36}=\frac{1}{6}\)

Card removed = king, queen and jack of clubs = 3 
Cards left = 52- 3 = 49
(i) Number. of hearts = 13
Probability of drawing a heart = \(\frac {13}{49}\) 
(ii) Number of king = 4
Number of kings left = 4 - 1 = 3
Probability of drawing a king = \(\frac {3}{49}\) 
(iii) 3 clubs are removed
Number of clubs left=13-3=10
Probability of drawing a club = \(\frac {10}{49}\)
(iv) There is only one '10' of hearts.
\(\therefore\)Probability of drawing a '10' of hearts =\(\frac {1}{49}\)

Total number of cards = 52
(i) Number of ace = 4
\(\therefore\)  Probability of drawing an ace = \(\frac{4}{52}=\frac{1}{13}\)
(ii) There is only one '2' of spades
\(\therefore\) Probability of drawing a '2' of spade = \(\frac{1}{52}\)
(iii) '10' of a black suit there are two cards
\(\therefore\)  Probability of drawing 10 of black suit = \(\frac{2}{52}=\frac{1}{26}\)

\(3\over26\)

Here, Total number of people=500
(i) Number of people who used car or only
=80+120=200
P (Number of people who used car or scooter only) = \(\frac{200} {500} = \frac{2} {5}\)
(ii) Number of who used only cycle = 110
P (Number of People who used only cycle) =\(\frac{110} {500} = \frac{11} {50}\) 
(iii) Number of people who used at least one kind of mode of transport 500 -20 = 480
P (Number of people who used at least one kind of mode of transport) = \(\frac{480} {500} = \frac{24} {25}\)
(iv) Keeping save environment good health and save energy in mind, cycle should be promoted as a mode of transport. 

\((a) \frac{81}{89} (b) \frac{8}{89}\)

Total number of cards = 48
Probability of an event = \(\frac{Total\ number \ of \ favourable \ outcomes}{Total\ number\ of \ outcomes}\)
Number of cards divisible by 7 = 7
P(cards divisible by 7) = \(\frac{7}{48}\)
Number of cards having a perfect square = 6
P(cards having a perfect square) =\(\frac{6}{48}\)=\(\frac{1}{8}\)
Number of multiples of 6 from 3 to 50 = 8
P(multiple of 6 from 3 to 50) = \(\frac{8}{48}\) =\(\frac{1}{6}\)

Since ΔADC is right angled at D.
\(\Rightarrow A C^{2}=A D^{2}+D C^{2}=6^{2}+8^{2}\) [by Pythagoras theorem)
= 36 + 64 = 100
\(\Rightarrow A C=10\) [taking positive sqaure root)
Now radius of circle = OC \(=\frac{1}{2} A C=5\)
P (dart will land in the shaded region)
\(=\frac{\text { Area of shaded region }}{\text { Area of the circle }} \)
\(=\frac{\text { Area of the circle }-\text { Area of the rectangle }}{\text { Area of the circle }} \)
\(= \frac{25 \pi-48}{25 \pi}\)

The possible outcomes of the experiment of tossing two coins are HH, HT, TH and TT.
(i) Outcomes favourable to Anu are HT and TH.
P (Anu getting first chance) = \(\frac{2}{4}=\frac{1}{2}\)
Outcomes favourable to Priya is HH.
\(\therefore \)  P ( Priya getting first chance) = \(\frac{1}{4}\)
Outcomes favourable to Jyoti is TT.
\(\therefore \)  P (Jyoti getting first chance) = \(\frac{1}{4}\)
(ii) No, the number of cases favourable to each one of them are not equal.
(iii) Dishonesty, as she kept two cases favourable for herself and one each for the other two friends.

(a) (ii) \(\frac{1}{8}\) 
(b) (i) \(\frac{1}{2}\) 
(c) (i) \(\frac{1}{2}\) 
(d) (iv) \(\frac{1}{4}\) 
(e) (iii) \(\frac{3}{4}\) 

Total number of footballs produced in one day = 220000
(i) (c): P(selecting a defective football) = \(\begin{equation} \frac{15}{100}=\frac{3}{20} \end{equation}\)
(ii) (d): P(selecting a non-defective football) = 1- P(selecting a defective football) = \(\begin{equation} 1-\frac{3}{20}=\frac{17}{20} \end{equation}\)
(iii) (b) : Total number of defective footballs produced in 1 day = \(\begin{equation} \frac{3}{20} \times 22000=3300 \end{equation}\)
(iv) (a) : Total number of non-defective footballs produced in 1 day = 22000 - 3300 = 18700
(v) (d): P(selecting defective football) \(\begin{equation} =\frac{8}{25} \end{equation}\)
P (selecting non-defective football) = \(\begin{equation} 1-\frac{8}{25}=\frac{17}{25} \end{equation}\)
Number of non-defective footballs produced \(\begin{equation} =\frac{17}{25} \times 22000=14960 \end{equation}\)