GATE CSE 2014 Set 3 | Question: 48

Question

Let $S$ be a sample space and two mutually exclusive events $A$ and $B$ be such that $A \cup B = S$. If $P(.)$ denotes the probability of the event, the maximum value of $P(A)P(B)$ is_____.

Comments

Use concept -->If perimeter is fix then square has maximum area. :)

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Use concept of arithmetic mean(AM) and geometric mean(GM)

for any two positive numbers

$AM\geq GM$

$AM = (P(A)+P(B))/2$

$GM = \sqrt{P(A)*P(B)}$

$(P(A)+P(B))/2 \geq \sqrt{P(A)*P(B)}$

Put $P(A)+P(B)=1$

So $P(A)*P(B)\leq 1/4$

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A U B = S → P(A U B) = P(S) = 1

P(A U B) = P(A) + P(B) – P(A ${\cap }^{}$ B)  [ P(A ${\cap }^{}$ B) = Phi,as they are mutually exclusive events]

So, P(AUB) = P(A) + P(B) = 1

P(A)xP(B) = P(A)x(1-P(A)) [ as P(A) + P(B) = 1]

Put, P(A) = Z , P(A)xP(B) = P(A)x(1-P(A)) = Zx(1-Z)

Maximum value of P(A)xP(B) = Maximum value of Zx(1-Z) = f(Z)

The maximum value of f(Z) will occur when f’(Z) =0  [f’(Z) =  Derivation of f(Z) with respect to Z]

Now f’(Z) = 1-2xZ = 0 → Z = 1/2

So, P(A)xP(B) = Zx(1-Z) will be maximum when Z = ½

So, Maximum value of P(A)xP(B) = Zx(1-Z) = (1/2)x(1-1/2) = ¼ = 0.25 

 

 

 

Answer 1

Let

P(A) = x, P(B) = 1-x  ($\because$ A and B are mutually exclusive and A $\cup$ B = S)

We want to find max value of

f(x) = P(A)*P(B) = x*(1-x) = x – $x^{2}$

we find critical points i.e $f^{'}(x)$ = 0

$f^{'}(x)$ = 1-2x = 0

$\therefore$ x = ½ , this x can be minima or maxima, so we will find second derivative and if  $f^{''}(x)$ < 0, then x will give max value for $f(x)$

$f^{''}(x)$ = -2 < 0

$\therefore$ at x = ½ we get max value

 =x(1-x) = ½ * ½ = ¼ = 0.25