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Five cards – the ten, jack, queen, king and ace of diamonds are well-shuffled with their faces downwards. One card is then picked up at random. 

(a) What is the probability that the drawn card is the queen?

I. 1/2 

II. 1/3 

III. 1/4 

IV. 1/5

(b) If the queen is drawn and put aside and a second card is drawn. What is the probability that the 2nd card is an ace? 

I. 1/2 

II. 1/3 

III. 1/4 

IV. 1/5 

(c) What is the probability that the second card is queen?

I. 1/2 

II. 1/3 

III. 1/4 

IV. None

(d) From five cards, two cards are drawn. Find the probability that the drawn cards are an ace and ten?

I. 1/5

II. 2/5 

III. 1/25 

IV. None

(e) The probability that an ordinary year has 53 Mondays is : 

I. 1/7 

II. 53/365 

III. 1/365 

IV. None

1 Answer

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Best answer

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.

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