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)

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

A 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

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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

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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)’"