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If A and B are two independent events such that $P(AB')=cfrac{3}{25}$ and $P(BA')=cfrac{8}{25}$ and $P(A)<P(B)$, then what is the value of $P(A)$?

I'm getting two answers of this question. So I can't be sure whether I am correct.Please help me solving this problem. Thanks in advance.

You can make a table to keep the overview:

$$begin{array}{c|c|c|c} & A&overline A & \ hline B & x & frac8{25} & y \ hline overline B& frac{3}{25}& & \ hline &z&& 1end{array}$$

If A and B are independent then the follwing to equations must hold:

$$P(B|A)=P(B)Rightarrow frac{z-frac3{25}}{z}=y$$

$$P(overline A|B)=P(overline A)Rightarrow frac{frac8{25}}{y}=1-z$$

Indeed this equation system has two solutions:

1) $y^*=frac25, z^*=frac15$

2) $y^*=frac45, z^*=frac35$

Choose the solution where $P(B)>P(A)Rightarrow y>z$

Edit:

If I´m right, both solutions fulfill the condition. Maybe this was your problem.

If $A, B$ are independent events then so are $A^{complement},B$ and $A,B^{complement}$ so we have $2$ equalities:

• $P(B)-P(A)P(B)=(1-P(A))P(B)=P(A^{complement})P(B)=P(A^{complement}cap B)=frac8{25}$




• $P(A)-P(A)P(B)=P(A)(1-P(B))=P(A)P(B^{complement})=P(Acap B^{complement})=frac3{25}$

Subtraction gives $P(B)-P(A)=frac5{25}=frac15$

Then substituting this in second equality gives: $P(A)-P(A)(frac15+P(A))=P=frac3{25}$

This leads two possibilities for $(P(A),P(B))$ but on base of $P(B)=frac15+P(A)>P(A)$ we can pick out the right one.

Use the fact that $P(B) = 1-P(B')$ and by independence $P(A cap B') = P(A) times P(B')$: You get two unknowns in two equations.