Describe the sample space for the following experiment.A coin is tossed four times.
A coin is tossed four times.
Sample space, S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT}
Given: A and B are two events.
P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35
By definition of P (A or B) under axiomatic approach we know that:
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
Now we have to find:
(i) P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
= 0.54 + 0.69 – 0.35
= 0.88
(ii) P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s Law}
P (A′ ∩ B′) = 1 – P (A ∪ B)
= 1 – 0.88
= 0.12
(iii) P (A ∩ B′) [This indicates only the part which is common with A and not B.
Hence this indicates only A]
P (only A) = P (A) – P (A ∩ B)
∴ P (A ∩ B′) = P (A) – P (A ∩ B)
= 0.54 – 0.35
= 0.19
(iv) P (A′ ∩ B) [This indicates only the part which is common with B and not A.
Hence this indicates only B]
P (only B) = P (B) – P (A ∩ B)
∴ P (A′ ∩ B) = P (B) – P (A ∩ B)
= 0.69 – 0.35
= 0.34
Answered by Sakshi | 1 month agoA and B are two events such that P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35. Find
P (A ∪ B)
Given:
P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35
By addition theorem, we have:P (A ∪ B) = P(A) + P (B) -P (A ∩ B) = 0.54 + 0.69 - 0.35
= 0.88
Concept: Event - Types of Events
Is there an error in this question or solution?
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A and B are two events such that P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35. Find
\[P (\bar{ A } \cap \bar{ B } )\]
Given:
P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35
\[P\left( \bar{A} \cap \bar{ B } \right) = 1 - P\left( A \cup B \right)\]
= 1 - 0.88
= 0.12
Concept: Event - Types of Events
Is there an error in this question or solution?
A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35.
Find
(i) P(A ∩ B)
(ii) P(A′ ∩ B′)
(iii) P(A ∩ B′) (iv) P(B ∩ A′)
It is given that P(A) = 0.54, P(B) = 0.69, P(A ∩ B) = 0.35
(i) We know that P (A ∪ B) = P(A) + P(B) − P(A ∩ B)
∴P (A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88
(ii) A′ ∩ B′ = (A ∪ B)′ [by De Morgan’s law]
∴P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) = 1 − 0.88 = 0.12
(iii) P(A ∩ B′) = P(A) − P(A ∩ B)
= 0.54 − 0.35
= 0.19
Concept: Event - Algebra of Events
Is there an error in this question or solution?
>
A and B are two events such that PA=0.54, PB=0.69 and PA ∩ B=0.35 . Find i PA ∩ B ii PA'∩ B' iii PA ∩ B' iv PB ∩ A'
Solution
Let, the two events Aand B such that,
P( A )=0.54 P( B )=0.69 P( A∩B )=0.35
(i)
The value of P( A∪B ) is calculated as follows,
P( A∪B )=P( A )+P( B )−P( A∩B ) P( A∪B )=0.54+0.69−0.35 P( A∪B )=0.88
(ii)
The value of P( A ′ ∩ B ′ ) is calculated as follows,
P( A ′ ∩ B ′ )=P ( A∪B ) ′ =1−P( A∪B ) =1−0.88 P( A ′ ∩ B ′ )=0.12
(iii) P( A∩ B ′ )
The value of P( A∩ B ′ )is calculated as follows,
P( A∩ B ′ )=P( A )−P( A∩B ) =0.54−0.35 P( A∩ B ′ )=0.19
(iv) P( B∩ A ′ )
The value of P( B∩ A ′ )is calculated as follows,
P( B∩ A ′ )=P( B )−P( A∩B ) =0.69−0.35 P( B∩ A ′ )=0.34
Mathematics
Math - NCERT
Standard XI
Last updated at Sept. 6, 2021 by
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Misc 7 A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (i) P(A ∪ B) Given P(A) = 0.54, P(B) = 0.69, P(A ∩ B) = 0.35 We know that P (A ∪ B) = P(A) + P(B) − P(A ∩ B) Putting values = 0.54 + 0.69 − 0.35 = 0.88
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Last updated at Aug. 28, 2021 by Teachoo
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Misc 7 A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (ii) P(A′ ∩ B′) P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) = 1 − 0.88 = 0.12 (By Demorgan law) Demorgan’s law "(A’"∩"B’) = (A "∪" B)’" "or (A’ "∪" B’) = (A "∩" B)’"