Question

Let A and B be two possible events of an experinment, such that P(A) = 0.4, P(B) = K and P(AUB) = 0.7. For what value of K are A and B independent?​

Answer 1

$\large\underline{\sf{Given- }}$

A and B be two possible events of an experinment, such that P(A) = 0.4, P(B) = K and P(AUB) = 0.7.

$\large\underline{\sf{To\:Find - }}$

The value of k for which event A and B are independent.

$\large\underline{\sf{Solution-}}$

Given that,

A and B be two possible events of an experinment, such that

↝ P(A) = 0.4

↝ P(B) = K

↝ P(AUB) = 0.7

↝ We know, By addition theorem,

$\boxed{ \bf{ \: P(A\cup B) = P(A) + P(B) - P(A\cap B)}}$

So, on substituting the values, we get

$\rm :\longmapsto\:0.7 = 0.4 + K - P(A\cap B)$

$\rm :\longmapsto\:0.7 - 0.4 - K = - \: P(A\cap B)$

$\rm :\longmapsto\:0.3 - K = - \: P(A\cap B)$

$\rm :\implies\: \boxed{ \rm \: P(A\cap B) = K - 0.3} - - - (1)$

Now,

↝ We know that, two events A and are independent iff

$\rm :\longmapsto\:P(A\cap B) = P(A) \times P(B)$

So, on substituting the values, we get

$\rm :\longmapsto\:K - 0.3 = 0.4 \times K$

$\rm :\longmapsto\:K - 0.3 = 0.4K$

$\rm :\longmapsto\:K - 0.4K = 0.3$

$\rm :\longmapsto\:0.6K = 0.3$

$\rm :\longmapsto\:K = \dfrac{0.3}{0.6}$

$\bf\implies \: \boxed{ \bf{ \: K \: = \: \dfrac{1}{2} }}$

Additional Information :-

$\boxed{ \bf{ \: P(A\cap B') = P(A) - P(A\cap B)}}$

$\boxed{ \bf{ \: P(B\cap A') = P(B) - P(A\cap B)}}$

$\boxed{ \bf{ \: P(B'\cap A') = 1 - P(A\cup B)}}$

$\boxed{ \bf{ \: P(B'\cup A') = 1 - P(A\cap B)}}$

Answer 2

Step-by-step explanation:

$</p><p>\large\underline{\sf{Given- }} </p><p>Given−</p><p> </p><p> </p><p></p><p>A and B be two possible events of an experinment, such that P(A) = 0.4, P(B) = K and P(AUB) = 0.7.</p><p></p><p>\large\underline{\sf{To\:Find - }} </p><p>ToFind−</p><p> </p><p> </p><p></p><p>The value of k for which event A and B are independent.</p><p></p><p>\large\underline{\sf{Solution-}} </p><p>Solution−</p><p> </p><p> </p><p></p><p>Given that,</p><p></p><p>A and B be two possible events of an experinment, such that</p><p></p><p>↝ P(A) = 0.4</p><p></p><p>↝ P(B) = K</p><p></p><p>↝ P(AUB) = 0.7</p><p></p><p>↝ We know, By addition theorem,</p><p></p><p>\boxed{ \bf{ \: P(A\cup B) = P(A) + P(B) - P(A\cap B)}} </p><p>P(A∪B)=P(A)+P(B)−P(A∩B)</p><p> </p><p> </p><p></p><p>So, on substituting the values, we get</p><p></p><p>\rm :\longmapsto\:0.7 = 0.4 + K - P(A\cap B):⟼0.7=0.4+K−P(A∩B)</p><p></p><p>\rm :\longmapsto\:0.7 - 0.4 - K = - \: P(A\cap B):⟼0.7−0.4−K=−P(A∩B)</p><p></p><p>\rm :\longmapsto\:0.3 - K = - \: P(A\cap B):⟼0.3−K=−P(A∩B)</p><p></p><p>\rm :\implies\: \boxed{ \rm \: P(A\cap B) = K - 0.3} - - - (1):⟹ </p><p>P(A∩B)=K−0.3</p><p> </p><p> −−−(1)</p><p></p><p>Now,</p><p></p><p>↝ We know that, two events A and are independent iff</p><p></p><p>\rm :\longmapsto\:P(A\cap B) = P(A) \times P(B):⟼P(A∩B)=P(A)×P(B)</p><p></p><p>So, on substituting the values, we get</p><p></p><p>\rm :\longmapsto\:K - 0.3 = 0.4 \times K:⟼K−0.3=0.4×K</p><p></p><p>\rm :\longmapsto\:K - 0.3 = 0.4K:⟼K−0.3=0.4K</p><p></p><p>\rm :\longmapsto\:K - 0.4K = 0.3:⟼K−0.4K=0.3</p><p></p><p>\rm :\longmapsto\:0.6K = 0.3:⟼0.6K=0.3</p><p></p><p>\rm :\longmapsto\:K = \dfrac{0.3}{0.6}:⟼K= </p><p>0.6</p><p>0.3</p><p> </p><p> </p><p></p><p>\bf\implies \: \boxed{ \bf{ \: K \: = \: \dfrac{1}{2} }}⟹ </p><p>K= </p><p>2</p><p>1</p><p> </p><p> </p><p> </p><p> </p><p></p><p>Additional Information :-</p><p>\boxed{ \bf{ \: P(A\cap B') = P(A) - P(A\cap B)}} </p><p>P(A∩B </p><p>′</p><p> )=P(A)−P(A∩B)</p><p> </p><p> </p><p></p><p>\boxed{ \bf{ \: P(B\cap A') = P(B) - P(A\cap B)}} </p><p>P(B∩A </p><p>′</p><p> )=P(B)−P(A∩B)</p><p> </p><p> </p><p></p><p>\boxed{ \bf{ \: P(B'\cap A') = 1 - P(A\cup B)}} </p><p>P(B </p><p>′</p><p> ∩A </p><p>′</p><p> )=1−P(A∪B)</p><p> </p><p> </p><p></p><p>\boxed{ \bf{ \: P(B'\cup A') = 1 - P(A\cap B)}} </p><p>P(B </p><p>′</p><p> ∪A </p><p>′</p><p> )=1−P(A∩B)</p><p> </p><p> </p><p>$