Question

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Answer 1

# i hope it will help you

Answer 2

$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,
$\underline{\textit{Statement:}}$
Product of Probabilities of happening and non-happening of an Event is $\frac{6}{25}$

$\underline{\large{\textsf{Step-1:}}}$
Assume a Variable for happening and non-happening of an Event

Let E represents the happening of an Event
and E' represents the non-happening of an Event

$\underline{\large{\textsf{Step-2:}}}$
Rewrite Statement-1

$\implies P(E) \times P(E') = \frac{6}{25}$

$\underline{\large{\textsf{Step-3:}}}$
Express the Probability of non- happening of Event in terms of Probability of happening of Event

$\implies P(E') = \frac{1}{P(E)} \times \frac{6}{25}$

$\implies P(E') = \frac{6}{25P(E)}$ ————[1]

$\underline{\large{\textsf{Step-4:}}}$
Using the Identity $\mathbf{P(S) = 1}$ (or $\textsf{Sum of Probabilities is equal to 1}$)

We have,
$S = E + E'$

And
$P(S) = P(E) + P(E')$

$\implies P(E) + P(E') = 1$ ————[2]

$\underline{\large{\textsf{Step-5:}}}$
Substituting [2] in [1]

$\implies P(E) + \frac{6}{25P(E)}= 1$

$\implies \frac{25(P(E))^{2} + 6}{25P(E)}= 1$

$\implies 25(P(E))^{2} + 6 =25P(E)$

$\implies 25(P(E))^{2} -25P(E) + 6 = 0$————[3]

$\underline{\large{\textsf{Step-5:}}}$
Solve for the Roots of Quadratic equation

We have,
$\textbf{For Quadratic equation ax^{2}+bx+c=0 }$

The Roots are
$\mathbf{x = \dfrac{-b+\sqrt{b^{2}-4ac}}{2a}}$,

$\mathbf{x = \dfrac{-b+\sqrt{b^{2}-4ac}}{2a}}$.

Comparing [3] with general form of Quadratic Equation

$a = 25, b = -25, c = 6$

On Substituting

One of the Root,
$P(E) = \dfrac{-(-25)+\sqrt{(-25)^{2}-(4(25)(6)}}{2(25)}$

$\implies P(E) = \dfrac{25+\sqrt{625-600}}{50}$

$\implies P(E) = \dfrac{25+\sqrt{25}}{50}$

$\implies P(E) = \dfrac{25+5}{50}$

$\implies P(E) = \dfrac{30}{50}$

$\implies P(E) = \dfrac{3}{5}$

And other Root,
$P(E) = \dfrac{-(-25)-\sqrt{(-25)^{2}-(4(25)(6)}}{2(25)}$

$\implies P(E) = \dfrac{25-\sqrt{625-600}}{50}$

$\implies P(E) = \dfrac{25-\sqrt{25}}{50}$

$\implies P(E) = \dfrac{25-5}{50}$

$\implies P(E) = \dfrac{20}{50}$

$\implies P(E) = \dfrac{2}{5}$

$\underline{\large{\textsf{Step-6:}}}$
Verify that Probabilities satisfy $\mathbf{0 \leq P(E) \leq 1}$

$P(E) = \dfrac{3}{5} = 0.6$

$0 \leq (0.6) \leq 1$($\textsf{Satisfied})$

$P(E) = \dfrac{2}{5} = 0.6$

$0 \leq (0.4) \leq 1$($\textsf{Satisfied})$

$\therefore$

$\textbf{Probability\: of\: happening\: of\: an\: Event =\underline{\mathbf{\dfrac{3}{5}\: or\: \dfrac{2}{5}}}}$