Below are given the probabilities A, B, C and D of an event. P(A) : $\dfrac{3}{5}$, P(B) : 1.2, P(C) : -1.2 and P(D) = $1\dfrac{2}{3}$. Which of the above values of the probabilities is possible :

1. Event A

2. Event B

3. Event C

4. Event D

A letter of English alphabet is chosen at random from English alphabets.

Assertion(A): The probability that the chosen letter is not a consonant is 5 : 52.

Reason(R): The probability of an event = $\dfrac{\text{Total number of outcomes}}{\text{Number of favourable outcomes}}$

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are false.

Face cards of spades are remove from the pack of 52 cards and the remaining cards are well shuffled. Then a card is drawn from the pack.

Statement (1): The probability of drawing a face card is $\dfrac{7}{52}$.

Statement (2): Kings, queens and jacks are the three face card and so the total number of face cards in the pack of 52 card is 3 x 4 = 12.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

In a lottery ticket, there are 20 prizes and 25 blanks.

Statement (1): Probability of not getting the prize = 1 - $\dfrac{20}{45}$

Statement (2): P(getting prize) + P(blank) = 1

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.