Two dice are thrown independently. Let A be the event that the number appeared on the $1^{\text {st }}$ die is less than the number appeared on the $2^{\text {nd }}$ die, B be the event that the number appeared on the $1^{\text {st }}$ die is even and that on the second die is odd, and C be the event that the number appeared on the $1^{\text {st }}$ die is odd and that on the $2^{\text {nd }}$ is even. Then

(A) the number of favourable cases of the event $(\mathrm{A} \cup \mathrm{B}) \cap \mathrm{C}$ is 6

(B) A and B are mutually exchusive

(C) The number of favourable cases of the events $\mathrm{A}, \mathrm{B}$ and C are 15, 6 and 6 respectively

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Probability

Two dice are thrown independently. Let A be the event that the number appeared on the

1^{st}

die is less than the number appeared on the

2^{nd}

die, B be the event that the number appeared on the

1^{st}

die is even and that on the second die is odd, and C be the event that the number appeared on the

1^{st}

die is odd and that on the

2^{nd}

is even. Then

(A)the number of favourable cases of the event $(A \cup B) \cap C$ is 6✓ Correct
(B)A and B are mutually exchusive
(C)The number of favourable cases of the events $A, B$ and C are 15, 6 and 6 respectively
(D)B and C are independent

Answer: (A)

For two dice, event

A

(first

<

second) has

5 + 4 + 3 + 2 + 1 = 15

favourable cases, while

B

(first even, second odd) and

C

(first odd, second even) each have

3 \times 3 = 9

. Now

(A \cup B) \cap C = (A \cap C) \cup (B \cap C)

; the part

B \cap C

is impossible because the first die cannot be simultaneously even and odd. The part

A \cap C

requires the first die odd, the second even and the first smaller, namely

(1, 2), (1, 4), (1, 6), (3, 4), (3, 6), (5, 6)

. Hence the number of favourable cases for

(A \cup B) \cap C

6

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option