Question

Question #16
The events A1 and A2 form the
partition of the sample space. If P(A1)
= P(A2) =1/2, P(
BA1) = 2/5, P(B|A2)
= 4/15 then P(A1B) =​

Answer 1

The events $A_1$ and $A_2$ form the partition of the sample space, which means,

• $A_1\cup A_2=S$

• $A_1\cap A_2=\phi$

That's why,

$\longrightarrow P(S)=P(A_1)+P(A_2)=1$

Given,

• $P(A_1)=P(A_2)=\dfrac{1}{2}$

• $P(B\,|\,A_1)=\dfrac{2}{5}$

• $P(B\,|\,A_2)=\dfrac{4}{15}$

We need to find $P(A_1\,|\,B)$

We see that,

$\longrightarrow B=B\cap S$

$\longrightarrow B=B\cap(A_1\cup A_2)$

$\longrightarrow B=(B\cap A_1)\cup(B\cap A_2)\quad\quad\dots(1)$

Also,

$\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap A_1\cap B\cap A_2$

$\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap A_1\cap A_2$

$\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap(A_1\cap A_2)$

$\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap\phi$

$\longrightarrow(B\cap A_1)\cap(B\cap A_2)=\phi\quad\quad\dots(2)$

(1) and (2) mean the events $B\cap A_1$ and $B\cap A_2$ form the partition of $B,$ hence,

$\longrightarrow P(B)=P(B\cap A_1)+P(B\cap A_2)\quad\quad\dots(3)$

But,

• $P(B\cap A_1)=P(B\,|\,A_1)\cdot P(A_1)$

• $P(B\cap A_2)=P(B\,|\,A_2)\cdot P(A_2)$

Then (3) becomes,

$\longrightarrow P(B)=P(B\,|\,A_1)\cdot P(A_1)+P(B\,|\,A_2)\cdot P(A_2)$

$\longrightarrow P(B)=\dfrac{2}{5}\cdot\dfrac{1}{2}+\dfrac{4}{15}\cdot\dfrac{1}{2}$

$\longrightarrow P(B)=\dfrac{1}{5}+\dfrac{2}{15}$

$\longrightarrow P(B)=\dfrac{1}{3}$

Then,

$\longrightarrow P(A_1\,|\,B)=\dfrac{P(A_1\cap B)}{P(B)}$

$\longrightarrow P(A_1\,|\,B)=\dfrac{P(B\cap A_1)}{P(B)}$

$\longrightarrow P(A_1\,|\,B)=\dfrac{P(B\,|\,A_1)\cdot P(A_1)}{P(B)}$

$\longrightarrow P(A_1\,|\,B)=\dfrac{\left(\dfrac{2}{5}\cdot\dfrac{1}{2}\right)}{\left(\dfrac{1}{3}\right)}$

$\longrightarrow\underline{\underline{P(A_1\,|\,B)=\dfrac{3}{5}}}$

Hence 3/5 is the answer.

Answer 2

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$

The events $A_1$ and $A_2$ form the partition of the sample space, which means,

$A_1\cup A_2=S$

$A_1\cap A_2=\phi$

That's why,

$\longrightarrow P(S)=P(A_1)+P(A_2)=1$

Given,

$P(A_1)=P(A_2)=\dfrac{1}{2}$

$P(B\,|\,A_1)=\dfrac{2}{5}$

$P(B\,|\,A_2)=\dfrac{4}{15}$

We need to find $P(A_1\,|\,B)$

We see that,

$\longrightarrow B=B\cap S$

$\longrightarrow B=B\cap(A_1\cup A_2)$

$\longrightarrow B=(B\cap A_1)\cup(B\cap A_2)\quad\quad\dots(1)$

Also,

$\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap A_1\cap B\cap A_2$

$\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap A_1\cap A_2$

$\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap(A_1\cap A_2)$

$\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap\phi$

$\longrightarrow(B\cap A_1)\cap(B\cap A_2)=\phi\quad\quad\dots(2)$

(1) and (2) mean the events $B\cap A_1$ and $B\cap A_2$ form the partition of $B,$ hence,

$\longrightarrow P(B)=P(B\cap A_1)+P(B\cap A_2)\quad\quad\dots(3)$

But,

$P(B\cap A_1)=P(B\,|\,A_1)\cdot P(A_1)$

$P(B\cap A_2)=P(B\,|\,A_2)\cdot P(A_2)$

Then (3) becomes,

$\longrightarrow P(B)=P(B\,|\,A_1)\cdot P(A_1)+P(B\,|\,A_2)\cdot P(A_2)$

$\longrightarrow P(B)=\dfrac{2}{5}\cdot\dfrac{1}{2}+\dfrac{4}{15}\cdot\dfrac{1}{2}$

$\longrightarrow P(B)=\dfrac{1}{5}+\dfrac{2}{15}$

$\longrightarrow P(B)=\dfrac{1}{3}$

Then,

$\longrightarrow P(A_1\,|\,B)=\dfrac{P(A_1\cap B)}{P(B)}$

$\longrightarrow P(A_1\,|\,B)=\dfrac{P(B\cap A_1)}{P(B)}$

$\longrightarrow P(A_1\,|\,B)=\dfrac{P(B\,|\,A_1)\cdot P(A_1)}{P(B)}$

$\longrightarrow P(A_1\,|\,B)=\dfrac{\left(\dfrac{2}{5}\cdot\dfrac{1}{2}\right)}{\left(\dfrac{1}{3}\right)}$

$\longrightarrow\underline{\underline{P(A_1\,|\,B)=\dfrac{3}{5}}}$

Hence 3/5 is the answer.