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One card is drawn at random from shuffled well deck of 52 cards. In which of the following cases are the events E and F independent?

(i)
E: the card drawn is a spade
F: the card drawn is a ace

(ii)
E: the card drawn is black
F: the card drawn is a king

(iii)
E: the card drawn is a king or queen
F: the card drawn is a jack

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In case I and ii

In case ii and (iii)

In case iii

In all cases

answer is A.

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Detailed Solution

Given,
Total cards = 52
Now, for two event A and B to be independent,

P (A \cap B) = p (A) \cdot P (B)

For the first case we have,
E: the card drawn is a spade
F: the card drawn is a ace
In the deck of card, 13 are spades and 4 are or space.
Total spade card So, the probability that card drawn is space

p (E) = \frac{26}{52} = \frac{1}{4}

The probability that card drawn is space,

p (F) = \frac{4}{52} = \frac{1}{13}

Here, only one card is ace of spade. So,

p (E \cap F) = \frac{1}{52}

Here,

(E \cap F) = p (E) \cdot p (F)

\frac{1}{52} = \frac{1}{4} \times \frac{1}{13}

\frac{1}{52} = \frac{1}{52}

Thus, E and F are independent for first case. For second case we hve,
E: the card drawn is black
F: the card drawn is a king

p (E) = \frac{13}{52} = \frac{1}{2} p (F) = \frac{4}{52} = \frac{1}{13}

P (E \cap F) = \frac{2}{52} = \frac{1}{26}

(E \cap F) = p (E) \cdot p (F)

\frac{1}{26} = \frac{1}{2} \times \frac{1}{13}

\frac{1}{26} = \frac{1}{26}

Thus, E and F are independent for second case.
For third case we have,
E: the card drawn is a king or queen
F: the card drawn is a jack
So,

\begin{matrix} p (E) = \frac{4}{52} + \frac{4}{52} = \frac{2}{13} \\ p (F) = \frac{4}{52} + \frac{4}{52} = \frac{2}{13} \\ p (E \cap F) = \frac{4}{52} \\ p (E \cap F) = p (E) \cdot p (F) \\ \frac{4}{52} \equiv \frac{2}{13} \times \frac{2}{13} \end{matrix}

\frac{1}{13} \neq \frac{4}{169}

Thus, E and F are not independent for third case.
Option 1 is correct.

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