Solveeit Logo
Login

Question

Question: Let \(E\) and \(F\) be two independent events. The probability that exactly one of them occurs is \(...

Let EE and FF be two independent events. The probability that exactly one of them occurs is 1125\dfrac{11}{25}​ and the probability of none of them occurring is 225\dfrac{2}{25}​. If P(T)P(T) denotes the probability of occurrence of the event TT then,
A. P(E)=45,P(F)=35P(E)=\dfrac{4}{5},P(F)=\dfrac{3}{5}
B. P(E)=15,P(F)=25P(E)=\dfrac{1}{5},P(F)=\dfrac{2}{5}
C. P(E)=25,P(F)=15P(E)=\dfrac{2}{5},P(F)=\dfrac{1}{5}
D. P(E)=35,P(F)=45P(E)=\dfrac{3}{5},P(F)=\dfrac{4}{5}

Explanation

Solution

Hint: We are given that P(E,F′)P(E,{{F}^{'}})or P(F,E′)=1125P(F,{{E}^{'}})=\dfrac{11}{25}. Consider Let P(E)=xP(E)=x and P(F)=yP(F)=y, then the probability that none of them occurring will be P(E′,F′)=225P({{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′)P(E,{{F}^{'}})or P(F,E′)=1125P(F,{{E}^{'}})=\dfrac{11}{25}.
So P(E,F′)+P(F,E′)=1125P(E,{{F}^{'}})+P(F,{{E}^{'}})=\dfrac{11}{25}.
Let P(E)=xP(E)=x and P(F)=yP(F)=y.
So, we can say that x(1−y)+y(1−x)=1125x(1-y)+y(1-x)=\dfrac{11}{25} ……… (1)
P(E′,F′)=225P({{E}^{'}},{{F}^{'}})=\dfrac{2}{25}
So, we can also say that (1−x)(1−y)=225(1-x)(1-y)=\dfrac{2}{25} ……. (2)
From (2) we get y=1−225(1−x)y=1-\dfrac{2}{25(1-x)}.
Substituting yy in (1) we get,

& 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{aligned}$$ Now solving the quadratic equation we get, $$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.