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.