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Q15(iii):
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ? (iii) E : ‘the card drawn is a king or queen’, F : ‘the card drawn is a queen or jack’.

Solution :

Step 1: Establishing the Sample Space

In a standard well-shuffled deck of cards, the total number of elementary outcomes in the sample space $S$ is $|S| = 52$. Each card has an equal probability of being drawn, defined as $P(\omega) = \frac{1}{52}$ for any $\omega \in S$.

Step 2: Defining Event E and Calculating its Probability

Event $E$: The card drawn is a King ($K$) or a Queen ($Q$).

  • Number of Kings = $4$
  • Number of Queens = $4$
  • $|E| = 4 + 4 = 8$

Using the probability definition $P(E) = \frac{|E|}{|S|}$ [By Classical Definition of Probability]:

$P(E) = \frac{8}{52} = \frac{2}{13}$

Step 3: Defining Event F and Calculating its Probability

Event $F$: The card drawn is a Queen ($Q$) or a Jack ($J$).

  • Number of Queens = $4$
  • Number of Jacks = $4$
  • $|F| = 4 + 4 = 8$

$P(F) = \frac{8}{52} = \frac{2}{13}$

Step 4: Intersection of E and F

$E \cap F$ represents the intersection of the sets: the card is (King or Queen) AND (Queen or Jack). The only common outcome is that the card is a Queen.

  • $E \cap F = \{ \text{Queen} \}$
  • $|E \cap F| = 4$ (There are 4 Queens in a deck)

$P(E \cap F) = \frac{4}{52} = \frac{1}{13}$

Step 5: Testing for Independence

Two events $E$ and $F$ are independent if and only if $P(E \cap F) = P(E) \times P(F)$ [By Definition of Stochastic Independence].

Calculate $P(E) \times P(F)$:

$P(E) \times P(F) = \left( \frac{2}{13} \right) \times \left( \frac{2}{13} \right) = \frac{4}{169}$

Compare with $P(E \cap F)$:

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

E F Q Venn Diagram: Intersection exists but condition fails

Analytical Conclusion

Since the product of the individual probabilities does not equal the probability of their intersection, the occurrence of $E$ alters the likelihood of $F$. Therefore, the events are dependent.

Final Solution: The events E and F are not independent, as $P(E \cap F) \neq P(E)P(F)$.


More Questions from Class 12 Mathematics Probability EXERCISE 13.2


CBSE Solutions for Class 12 Mathematics Probability


Chapters in CBSE - Class 12 Mathematics


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