We have five cards – ten, jack, queen, king and ace of diamonds. 

Total cards = n(S) = 5. 

(a) Option (iv)

Total queen cards in given set of cards n(E₁) = 1. 

∴ Probability that drawn card is a queen = \(\frac{n(E_1)}{ n(S)}\) = \(\frac{1}{5}\). 

Hence,

Option (iv) is correct. 

(b) Option (iii)

If the queen is drawn and put aside. 

Then total number of remaining cards = n(S') = 4. 

And total ace cards in remaining set of cards = n(E₁) = 1. 

∴ The probability that 2nd card is an ace = \(\frac{n(E_1)}{ n(S')}\) = \(\frac{1}{4}\) . 

Hence, 

Option (iii) is correct. 

(c) Option (iv) 

Total queen cards in the remaining set of cards = n(E₃) = 0.

∴ The probability that 2nd card is queen card = \(\frac{n(E_3)}{ n(S') }\)

= \(\frac{0}{4}\) 

= 0. 

Hence,

Option (iv) is correct. 

(d) Option (iv)

From five cards, 

Two cards are drawn. 

Let \(E'_4\) = Event that first card is ten and second card is ace card. 

\(E''_4\) = Event that first card is ace card and second card is ten card. 

S₄ = total possible outcomes in drawing two cards = 5 × 4 = 20. 

∴ Probability that drawn card are an ace card and ten card = P(\(E'_4\)) + P(\(E''_4\)) 

= \(\frac{1}{5\times 4}\) + \(\frac{1}{5\times 4}\) 

= \(\frac{2}{20}\) 

= \(\frac{1}{10}\)

Hence,

The correct answer is option (iv). 

(e) Option (i) 

In an ordinary year, total day = 365 

= 52 weeks and 1 day. 

Hence,

There will be 52 Mondays for sure. 

So,

In ordinary year there will be 52 Mondays and 1 day will be left This one day could be a Monday or Tuesday or Wednesday or Thursday or Friday or Saturday or Sunday. 

Hence,

Total possible outcomes = 7 

But favorable outcome = 1. 

∴ Probability of getting 53 Mondays = \(\frac{1}{7}\). 

Hence, 

Option (i) is correct.