22. 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′
22. 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′
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1 Answer
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22. The sample space of the experiment is
S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Now, A = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
B = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(3, 1), (3, 2), (3, 3), (3, 4
...more
Similar Questions for you
3, 4, 5, 5
In remaining six places you have to arrange
3, 4, 5,5
So no. of ways
Total no. of seven digits nos. =
Hence Req. prob.
f (x) = x? – 4x + 1 = 0
f' (x) = 4x³ – 4
= 4 (x–1) (x²+1+x)
=> Two solution
Let z be equal to (x + iy)
(x + iy) + (x – iy) = (x + iy)2 (i + 1)
Equating the real & in eg part.
(i) & (ii)
4xy = -2x Þ x = 0 or y =
(for x = 0, y = 0)
For y =
x2
x =
=
of
=
When
gives c = 1
So
sum of all solutions =
Hence k = 42
Each element of ordered pair (i, j) is either present in A or in B.
So, A + B = Sum of all elements of all ordered pairs {i, j} for and
= 20 (1 + 2 + 3 + … + 10) = 1100
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