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Given,
P(A) = 0.4
P(B) = 0.5
∴ A and B are mutually exclusive events, then P(A ∩ B) = 0
Now,
(i) P(A ∩ B) = P(A) + P(B)
= 0.4 + 0.5
= 0.9
∴ P(A ∪ B) = 0.9
= 1 - 0.9
= 0.1
= 0.5 - 0
= 0.4 - 0
= 0.4
P(A) = 0.54
P(B) = 0.69
P(A ∩ B) = 0.35
(i) P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
= 0.54 + 0.69 - 0.35
= 1.23 - 0.35
∴ P(A ∪ B) = 0.88
= 1 - 0.88
= 0.12
= 0.54 - 0.35
= 0.19
= 0.69 - 0.35
= 0.34
(i) Given,
(ii) Given,
P(A) = 0.35, P(B) = ...
P(A ∩ B) = 0.25, P(A ∪ B) = 0.6
∵ P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
0.6 = 0.35 + P(B) - 0.25
0.6 = 0.10 + P(B)
P(B) = 0.6 - 0.1
(iii) Given,
P(A) = 0.5, P(B) = 0.35
P(A ∩ B) =...., P(A ∪ B) = 0.7
0.7 = 0.5 + 0.35 - P(A ∩ B)
0.7 = 0.85 - P(A ∩ B)
P(A ∩ B) = 0.85 - 0.7
P(A ∩ B) = 0.15
We know by addition theorem on probability
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
⟹ 0.5 = 0.3 + 0.4 - P(A ∩ B)
P(A ∩ B) = 0.3 + 0.4 - 0.5
= 0.7 - 0.5
= 0.2
∴ P(A ∩ B) = 0.2
= 0.5 + 0.3 - 0.2
= 0.8 - 0.2
= 0.6
∴ P(A ∪ B) = 0.6
We know,
P(A ∪ B) = 0.8
P(A ∩ B) = 0.3
⟹ 1 - P(A) = 0.5
⟹ P(A) = 1 - 0.5 = 0.5
Now, by addition theorem on probabiltiy
0.8 = 0.5 + P(B) - 0.3
0.8 = P(B) + 0.2
P(B) = 0.8 - 0.2
∴ P(B) = 0.6
P(A ∪ B) = P(A) + P(B)
∵ A, B and C are mutually exhaustive
∴ A ∪ B ∪ C = S ⟹ P(A ∪ B ∪ C) = P(S)
⟹ P(A) + P(B) + P(C) = 1
P(C) = 1 – {P(A) + P(B)}
∴ odds against C is
= 43 : 34
Let chance in favour of other be x
Odds in favour of other
∵ I card is drawn from a well shuffled deck of 52 cards
∴ S = 52C1 = 52
The favourable events is that drawn card is either spade or a king
Let A = Event of choosing shade
⟹ 13C1 = 13
B = Event of choosing a king
⟹ 4C1 = 4
Also, king can be of spade
∴ (A ∩ B) = 1
∴ P(A ∩ B) = P(A) + P(B) - P(A ∩ B)
Since two dice is thrown,
∴ S = 62 = 36
Let A be the event of choosing doublet
= {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
B the event of choosing total of 9.
{(3, 6), (4, 5), (5, 4), (6, 3)}
∴ Probability of choosing neither a doublet nor a total of 9.
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Chapter 33: Probability – Exercise 33.1...
Chapter 33: Probability – Exercise 33.2...
Chapter 33: Probability – Exercise 33.3...