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Two independent events $\mathrm{A}$ and $\mathrm{B}$ have $\mathrm{P}(\mathrm{A})=\frac{1}{3}$
and $\mathrm{P}(\mathrm{B})=\frac{3}{4}$. What is the probability that exactly one of the two events A or B occurs?
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and $\mathrm{P}(\mathrm{B})=\frac{3}{4}$. What is the probability that exactly one of the two events A or B occurs?
Solution:
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Verified Answer
The correct answer is:
$\frac{7}{12}$
$A$ and $B$ are independent.
$P(A)=\frac{1}{3}$
$P(B)=\frac{3}{4}$
We want to find probability that exactly one of the two events $A$ or $B$ occurs i.e. when $A$ occurs $B$ does not and vice-versa. Lets take desired prob. is $P$. $\therefore P=P(A)(1-P(B))+P(B)(1-P(A))$
$$
\begin{array}{l}
=\frac{1}{3}\left(1-\frac{3}{4}\right)+\frac{3}{4}\left(1-\frac{1}{3}\right) \\
=\frac{1}{3} \times \frac{1}{4}+\frac{3 \times 2}{12} \\
P=\frac{7}{12}
\end{array}
$$
$P(A)=\frac{1}{3}$
$P(B)=\frac{3}{4}$
We want to find probability that exactly one of the two events $A$ or $B$ occurs i.e. when $A$ occurs $B$ does not and vice-versa. Lets take desired prob. is $P$. $\therefore P=P(A)(1-P(B))+P(B)(1-P(A))$
$$
\begin{array}{l}
=\frac{1}{3}\left(1-\frac{3}{4}\right)+\frac{3}{4}\left(1-\frac{1}{3}\right) \\
=\frac{1}{3} \times \frac{1}{4}+\frac{3 \times 2}{12} \\
P=\frac{7}{12}
\end{array}
$$
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