26. A die is thrown, find the probability of following events:
(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.
26. A die is thrown, find the probability of following events:
(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.
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1 Answer
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26. The sample space of throwing s dice is
S = {1, 2, 3, 4, 5, 6}, n (S) = 6.
(i) Let A be event such that a prime number will appear. Then,
A = {2, 3, 5}
? n (A) = 3
Here; P (A) =
(ii) Let B be event such that a number greater than or equal to 3 will appear. Then
B = {3, 4, 5, 6}
So, n (B) = 4
Therefore P (B) =
(iii) Let C be event such that a number less than or equal to one will appear. Then,
C = {1}
So, n (C) = 1
? P (C) =
(iv) Let D be event such that a number more than 6 appears. Then,
D =∅
So, n (D) = 0
? P (D) =
(v) Let E be event such that a number less than 6 appears. Then
E = {1, 2, 3, 4,
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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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