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Question : 147 of 150
Marks:
+1 ,
-0
Solution:
‌ ‌ Given, ‌ P ( A ∩ B ) = ‌ ‌ ⇒ ‌ ‌ P ( A ) ⋅ P ( B ) = ‌ (
∵ A and
B are independent events)
and
‌ ‌ P ( A ∩ B ) = ‌ ⇒ P ( A ∪ B ) = ‌ ⇒ 1 − P ( A ∪ B ) = ‌ ⇒ P ( A ∪ B ) = 1 − ‌ = ‌ ⇒ P ( A ) + P ( B ) − P ( A ∩ B ) = ‌ ⇒ P ( A ) + P ( B ) − P ( A ) ⋅ P ( B ) = ‌ ⇒ P ( A ) + P ( B ) = ‌ + ‌ = ‌ = ‌ Now,
P ( A ) − P ( B ) = √ { ( P ( A ) + P ( B ) } 2 − 4 P ( A ) ⋅ P ( B ) . = √ ( ‌ ) 2 − 4 ( ‌ ) ‌ ‌ ffrom Eqs. (i) and (ii)]
‌ ‌ = √ ‌ − ‌ = √ ‌ = ‌ ‌ ⇒ ‌ ‌ P ( A ) − P ( B ) = ‌ On adding Eqs. (ii) and (iii), we get
2 P ( A ) ‌ = ‌ + ‌ = 1 ⇒ ‌ ‌ P ( A ) ‌ = ‌ On putting the value of
P ( A ) in Eq. (iii), we get
‌ ‌ − P ( B ) ‌ = ‌ ⇒ ‌ ‌ P ( B ) = ‌ − ‌ ‌ = ‌ = ‌ ⇒ P ( B ) ‌ = ‌ ‌ Hence, ‌ ‌ ‌ P ( A ) ‌ = ‌ ‌ and ‌ P ( B ) = ‌
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