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Exercise 14.1 · Q6

Q.Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice ≤5\leq 5. Describe the events

(i) A′A'
(ii) not B
(iii) A or B
(iv) A and B
(v) A but not C
(vi) B or C
(vii) B and C
(viii) A∩B′∩C′A \cap B' \cap C'.
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Two dice give a sample space of 3636 ordered pairs. With AA: even on the first die, BB: odd on the first die, CC: sum ≤5\le 5, we get (i) A′=BA'=B,

(ii) not B=AB=A,

(iii) A∪B=SA\cup B=S,

(iv) A∩B=∅A\cap B=\varnothing,

(v) A∩C′A\cap C' has 1414 outcomes,

(vi) B∪CB\cup C has 2222 outcomes,

(vii) B∩CB\cap C has 66 outcomes,

(viii) A∩B′∩C′=A∩C′A\cap B'\cap C'=A\cap C' has 1414 outcomes.

When two dice are thrown, each outcome is an ordered pair (i,j)(i,j) with ii from the first die and jj from the second. There are 6×6=366\times 6=36 equally likely outcomes.

The three given events

AA — even number on the first die (i∈{2,4,6}i\in\{2,4,6\}): ∣A∣=18|A|=18.

BB — odd number on the first die (i∈{1,3,5}i\in\{1,3,5\}): ∣B∣=18|B|=18.

CC — sum ≤5\le 5:

C={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)},∣C∣=10.C=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)\},\qquad |C|=10.

Describing each requested event

(i) A′A' — "not even on the first die" means odd on the first die, so A′=BA'=B (an odd number on the first die).

(ii) not BB — "not odd on the first die" means even on the first die, so B′=AB'=A.

(iii) AA or BB =A∪B=A\cup B — the first die is even or odd, which is always the case, so A∪B=SA\cup B=S (all 3636 outcomes).

(iv) AA and BB =A∩B=A\cap B — the first die cannot be even and odd at once, so A∩B=∅A\cap B=\varnothing.

(v) AA but not CC =A∩C′=A\cap C' — even first die and sum >5>5. Remove from AA the outcomes with sum ≤5\le 5, namely (2,1),(2,2),(2,3),(4,1)(2,1),(2,2),(2,3),(4,1):

A∩C′={(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)},A\cap C'=\{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\},

which has 18−4=1418-4=14 outcomes. …

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