Question
Three events $A, B$ and $C$ in a sample space are mutually exclusive and exhaustive. If $P(C) = 0.8$ and $3P(B) = 2P(A’),$ then find $P(A)$ and vP(B).$
✓
Answer
Here, $P(C’) = 0.8, 3P(B) = 2P(A’)$ are given.
$P(C’) = 0.8 \therefore P(C) = 1 – P(C’) = 1 – 0.8 = 0.2$
$3P(B) = 2P(A)$
$\therefore 3 P(B)=2[1-P(A)]$
$\therefore 3 P(B)=2-2 P(A)$
$\therefore P(B)=\frac{2-2 P(A)}{3}$
$A, B$ and $C$ are mutually exclusive and exhaustive events.
$\therefore P(A \cup B \cup C)=1 .$
$\therefore P(A)+P(B)+P(C)=1$
$\therefore P(A)+\frac{2-2 P(A)}{3}+0.2=1$
$\therefore \frac{3 P(A)+2-2 P(A)}{3}=1-0.2$
$\therefore \frac{P(A)+2}{3}=0.8$
$\therefore P(A)+2=0.8 \times 3=2.4$
$\therefore P(A)=2.4-2=0.4$
Now, putting $P(A)=0.4$ in $P(B)=\frac{2-2 P(A)}{3}$,
$P(B)=\frac{2-2(0.4)}{3}=\frac{2-0.8}{3}=\frac{1.2}{3}=0.4$
$P(C’) = 0.8 \therefore P(C) = 1 – P(C’) = 1 – 0.8 = 0.2$
$3P(B) = 2P(A)$
$\therefore 3 P(B)=2[1-P(A)]$
$\therefore 3 P(B)=2-2 P(A)$
$\therefore P(B)=\frac{2-2 P(A)}{3}$
$A, B$ and $C$ are mutually exclusive and exhaustive events.
$\therefore P(A \cup B \cup C)=1 .$
$\therefore P(A)+P(B)+P(C)=1$
$\therefore P(A)+\frac{2-2 P(A)}{3}+0.2=1$
$\therefore \frac{3 P(A)+2-2 P(A)}{3}=1-0.2$
$\therefore \frac{P(A)+2}{3}=0.8$
$\therefore P(A)+2=0.8 \times 3=2.4$
$\therefore P(A)=2.4-2=0.4$
Now, putting $P(A)=0.4$ in $P(B)=\frac{2-2 P(A)}{3}$,
$P(B)=\frac{2-2(0.4)}{3}=\frac{2-0.8}{3}=\frac{1.2}{3}=0.4$
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