Question

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. Describe the events.

(i) 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'​

Answer 1

Step-by-step explanation:

The Sample Space S for this random experiment will have 36 possible outcomes. {As they are 6 on one dice and another 6 on other dice, So, totally, 6 * 6 = 36}

Now, A is an event given that even number on the first die

2,4 and 6 are the numbers which can come on first die to satisfy the event

And for each of them we have 6 possibilities

For eg, 2 has {2,1},{2,2}, {2,3},{2,4},{2,5}{2,6}. Similarly 6 for 4 and 6

So, totally we have 3 * 6 = 18 chances for event A

Now, B is an event given that odd number on the first die

1,3 and 5 are the numbers which can come on first die to satisfy the event

And for each of them we have 6 possibilities

For eg, 1 has {1,1},{1,2}, {1,3},{1,4},{1,5}{1,6}. Similarly 6 for 3 and 5

So, totally we have 3 * 6 = 18 chances for event B

Now, C is an given that the sum of the numbers on the dice ≤ 5.

The possibilities are (1,1), (1,2),(1,3), (1,4), (2,1),(2,2),(2,3), (3,1), (3,2), (4,1)

n(C) = 10

Coming back to the question,

(i) A' = not Getting an even number on the first die which means the event (B)

So, n(A') = 18

(ii) not B = B' = not Getting an odd number on the first die. = A

n(B') = 18

(iii) A or B = A ∪ B = S (Sample space)

n(A ∪ B) = 36

(iv) A and B = A ∩ B

n(A∩B) = 0 Intersection of A and B is 0

(v) A but not C = A ∩ C'

So, intersection of A and C are  (1,1), (1,2),(1,3), (1,4),(3,1), (3,2)

So, n(A ∩ C) = 6

But, n(A ∩ C') = n(A) - n(A ∩ C) = 18 - 6 = 12

n(A ∩ C') = 12

(vi) B or C = B ∪ C

Terms which are in C but not in B are = n(C) - n ((A ∩ C)') = 10 - 4 = 6

So, n(B ∪ C) = n(B) + n(C) - n ((A ∩ C)') = 18 + 6 = 24

n(B ∪ C) = 24

(vii) B and C = B ∩ C

Only 2 of them match with B and C

So, n(B ∩ C) = 2

(viii) A ∩ B' ∩ C'​ =

A ∩ B' means A but not B which is finally A

now A and C'​ is A ∩ C' which is same as (v) condition

n(A ∩ B' ∩ C') = n (A ∩ C'​) = 12

Hope it helps!

{I gave a explanation in ur another account[SorkoZom](in one math question), please read it.}