Question

If A and B are two events and P (A) = 2/3, P(B) =3/5, P(AUB) = 5/6, then the value of P(A¹/B¹) =​

Answer 1

⇝ Given :-

• A and B are two events

• P (A) = $\sf \dfrac{2}{3}$

• P(B) = $\sf \dfrac{3}{5}$

• P(AUB) = $\sf \dfrac{5}{6}$

⇝ To Find :-

• Value of $\sf P\bigg(\dfrac{A^1}{B^1}\bigg)$

⇝ Solution :-

As P (B) = $\sf \dfrac{3}{5}$

$:\longmapsto \sf P(B{}^{1} ) = 1 -P(B)$

$:\longmapsto \sf P(B {}^{1} ) = 1 - \frac{3}{5}$

$:\longmapsto \bf P(B {}^{1} ) = \frac{2}{5} \: - - - (1)$

As P (AUB) = $\sf \dfrac{5}{6}$

$:\longmapsto \sf P((AUB) {}^{1} ) = 1 -P(AUB)$

$:\longmapsto \sf P((AUB) {}^{1} ) = 1 - \frac{5}{6}$

$:\longmapsto \bf P((AUB) {}^{1} ) = \frac{1}{6} \: - - - (2)$

❒ We Know By De - Morgan's Law :

$\bf\orange{{P((AUB)} {}^{1} ) = P(A {}^{1}\pmb{∩}B{}^{1} )}$

$:\longmapsto \bf P(A {}^{1}\pmb{∩}B{}^{1} ) = \frac{1}{6} \: ---(3)$

We Know :

$\bf P \bigg( \frac{X}{Y} \bigg) = \frac{P(X\pmb{∩}Y)}{P(Y)}$

Hence,

$\sf P\bigg(\dfrac{A^1}{B^1}\bigg) = \dfrac{P(A^1\pmb{∩}B^1)}{P(B^1)}$

⏩ Using (1) and (3) :

$:\longmapsto \sf P\bigg(\dfrac{A^1}{B^1}\bigg) = = \frac{1/6}{2/5}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf P\bigg(\dfrac{A^1}{B^1}\bigg) = \frac{5}{12} } }}}$

Answer 2

$\huge \rm {\underline{\underbrace{Elucidation:-}}}$

➻This Question is based on 'De-morgan's law'

➻It states that The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements.

_________________________

➻Comming to the problem,

$\sf \red {\underline{\underline{provided\: that:}}}$

$\to \tt {A\:and\:B\:are\:two\:events}$

$\to \tt {P (A) = \frac{2}{3}}$

$\to \tt {P (B) = \frac{3}{5}}$

$\to \tt { (A∪B)= \frac{5}{6}}$

________________________

$\sf \blue {\underline{\underline{To\: be\: Determined:}}}$

$\large \to \tt { p{(\frac{A^{1}}{B^{1}})}=?}$

➻It is known to be written as,

$\large \to \tt \green{ p{(\frac{A^{1}}{B^{1}})}=\frac{P(A^{1}∩ B^{1})}{P(B^{1})}}$--->Equation-1

${ [∵p{(\frac{A}{B})}=\frac{P(A∩ B)}{P(B)}]}$

________________________

$\sf \pink {\underline{\underline{We\: Know:}}}$

$\to \tt {P(B^{1})=1-P(B)}$

${[*P(B)=\frac{3}{5}*]}$

$\large \to \tt {1-\frac{3}{5}}$

➻By taking L.C.M,

$\large \to \tt {\frac{5-3}{5}}$

$\large \to \tt \green{P(B^{1})=\frac{2}{5}}$---> Equation-2

________________________

$\sf \purple {\underline{\underline{Similarly,}}}$

$\to \tt {P(A∪B^{1})=1-P(A∪B)}$

${[*P(A∪B)=\frac{5}{6}*]}$

$\large \to \tt {1-\frac{5}{6}}$

By Taking LCM,

$\large \to \tt {\frac{6-5}{6}}$

$\large \to \tt \green {P(A∪B^{1})=\frac{1}{6}}$ --->Equation-3

________________________

➻According to De Morgan’s first law,

➻The complement of the union of two sets A and B is equal to the intersection of the complement of the sets A and B.

$\implies \tt {P(A∪B)^{1}= P(A^{1}∩ B^{1})}$

________________________

➻From Equation-3,

$\large \to \tt \green {P(A^{1}∩ B^{1})=\frac{1}{6}}$ ----->Equation-4

➻From Equation-1 , Equation-2& Equation-4 we can conclude that,

$\large \to \tt { p{(\frac{A^{1}}{B^{1}})}=\frac{P(A^{1}∩B^{1})}{P(B^{1})}}$

$\large \to \tt { p{(\frac{A^{1}}{B^{1}})}=\frac{\frac{1}{6}}{\frac{2}{5}}}$

$\large \to \tt {p{(\frac{A^{1}}{B^{1}})}=\frac{1\times5}{6\times2}}$

$\large \colon \implies \green {p{(\frac{A^{1}}{B^{1}})}=\frac{5}{12}}$

________________________

$\sf \red {\underline{\underline{Henceforth,}}}$

➻The required answer is, $\bold {\frac{5}{12}}$