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Question

Answers

A. $P(E)=\dfrac{4}{5},P(F)=\dfrac{3}{5}$

B. $P(E)=\dfrac{1}{5},P(F)=\dfrac{2}{5}$

C. $P(E)=\dfrac{2}{5},P(F)=\dfrac{1}{5}$

D. $P(E)=\dfrac{3}{5},P(F)=\dfrac{4}{5}$

Answer
Verified

Hint: We are given that $P(E,{{F}^{'}})$or $P(F,{{E}^{'}})=\dfrac{11}{25}$. Consider Let $P(E)=x$ and $P(F)=y$, then the probability that none of them occurring will be $P({{E}^{'}},{{F}^{'}})=\dfrac{2}{25}$. Solve it, you will get the answer.

Complete step-by-step answer:

It is given in question that $P(E,{{F}^{'}})$or $P(F,{{E}^{'}})=\dfrac{11}{25}$.

So $P(E,{{F}^{'}})+P(F,{{E}^{'}})=\dfrac{11}{25}$.

Let $P(E)=x$ and $P(F)=y$.

So, we can say that $x(1-y)+y(1-x)=\dfrac{11}{25}$ â€¦â€¦â€¦ (1)

$P({{E}^{'}},{{F}^{'}})=\dfrac{2}{25}$

So, we can also say that $(1-x)(1-y)=\dfrac{2}{25}$ â€¦â€¦. (2)

From (2) we get $y=1-\dfrac{2}{25(1-x)}$.

Substituting $y$ in (1) we get,

\[\begin{align}

& x(1-1+\dfrac{2}{25(1-x)})+\left( 1-\dfrac{2}{25(1-x)} \right)(1-x)=\dfrac{11}{25} \\

& \dfrac{2x}{25(1-x)}+\left( (1-x)-\dfrac{2}{25} \right)=\dfrac{11}{25} \\

& \dfrac{2x}{25(1-x)}+(1-x)=\dfrac{13}{25} \\

& 2x+25{{(1-x)}^{2}}=13(1-x) \\

& 25(1-2x+{{x}^{2}})=13-15x \\

& 25{{x}^{2}}-35x+12=0 \\

\end{align}\]

Now solving the quadratic equation we get,

\[25{{x}^{2}}-35x+12=0\]

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{35\pm \sqrt{{{35}^{2}}-4(25)(12)}}{2(25)}=\dfrac{35\pm \sqrt{25}}{2(25)}=\dfrac{35\pm 5}{2(25)}$

So we get, $x=\dfrac{3}{5},\dfrac{4}{5}$. Now substituting these two values of $x$ in the above equation, $y=1-\dfrac{2}{25(1-x)}$, we get, $y=1-\dfrac{2}{25(1-\dfrac{4}{5})}=\dfrac{3}{5}$ and $y=1-\dfrac{2}{25(1-\dfrac{3}{5})}=\dfrac{4}{5}$.

So we get, \[y=\dfrac{3}{5},\dfrac{4}{5}\].

Therefore, $x=\dfrac{4}{5},\dfrac{3}{5}$ and \[y=\dfrac{3}{5},\dfrac{4}{5}\].

Hence, the cases possible are : $P(E)=\dfrac{4}{5},P(F)=\dfrac{3}{5}$ and $P(E)=\dfrac{3}{5},P(F)=\dfrac{4}{5}$.

So the correct answers are options (A) and (D).

Note: Read the question carefully. Your concept regarding independent events should be clear. The occurrence of event E will not affect the occurrence of event F. An example of the same could be tossing a coin and rolling a die at the same time.

Complete step-by-step answer:

It is given in question that $P(E,{{F}^{'}})$or $P(F,{{E}^{'}})=\dfrac{11}{25}$.

So $P(E,{{F}^{'}})+P(F,{{E}^{'}})=\dfrac{11}{25}$.

Let $P(E)=x$ and $P(F)=y$.

So, we can say that $x(1-y)+y(1-x)=\dfrac{11}{25}$ â€¦â€¦â€¦ (1)

$P({{E}^{'}},{{F}^{'}})=\dfrac{2}{25}$

So, we can also say that $(1-x)(1-y)=\dfrac{2}{25}$ â€¦â€¦. (2)

From (2) we get $y=1-\dfrac{2}{25(1-x)}$.

Substituting $y$ in (1) we get,

\[\begin{align}

& x(1-1+\dfrac{2}{25(1-x)})+\left( 1-\dfrac{2}{25(1-x)} \right)(1-x)=\dfrac{11}{25} \\

& \dfrac{2x}{25(1-x)}+\left( (1-x)-\dfrac{2}{25} \right)=\dfrac{11}{25} \\

& \dfrac{2x}{25(1-x)}+(1-x)=\dfrac{13}{25} \\

& 2x+25{{(1-x)}^{2}}=13(1-x) \\

& 25(1-2x+{{x}^{2}})=13-15x \\

& 25{{x}^{2}}-35x+12=0 \\

\end{align}\]

Now solving the quadratic equation we get,

\[25{{x}^{2}}-35x+12=0\]

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{35\pm \sqrt{{{35}^{2}}-4(25)(12)}}{2(25)}=\dfrac{35\pm \sqrt{25}}{2(25)}=\dfrac{35\pm 5}{2(25)}$

So we get, $x=\dfrac{3}{5},\dfrac{4}{5}$. Now substituting these two values of $x$ in the above equation, $y=1-\dfrac{2}{25(1-x)}$, we get, $y=1-\dfrac{2}{25(1-\dfrac{4}{5})}=\dfrac{3}{5}$ and $y=1-\dfrac{2}{25(1-\dfrac{3}{5})}=\dfrac{4}{5}$.

So we get, \[y=\dfrac{3}{5},\dfrac{4}{5}\].

Therefore, $x=\dfrac{4}{5},\dfrac{3}{5}$ and \[y=\dfrac{3}{5},\dfrac{4}{5}\].

Hence, the cases possible are : $P(E)=\dfrac{4}{5},P(F)=\dfrac{3}{5}$ and $P(E)=\dfrac{3}{5},P(F)=\dfrac{4}{5}$.

So the correct answers are options (A) and (D).

Note: Read the question carefully. Your concept regarding independent events should be clear. The occurrence of event E will not affect the occurrence of event F. An example of the same could be tossing a coin and rolling a die at the same time.

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