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$\mathrm{A}$ student appears for tests 1 , 11 and $\mathrm{III}$. The student is successful if he passes in tests I,II or I, III. The probabilities of the student passing in tests I. II and III are respectively,
p,q and $1 / 2 .$ If the probability of the student to be successful is $1 / 2 .$ Then
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p,q and $1 / 2 .$ If the probability of the student to be successful is $1 / 2 .$ Then
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Verified Answer
The correct answer is:
$p(1+q)=1$
Let $X$ be the event that student will be successful. $X_{1}$ be the event that student will be pass in test-I. $X_{2}$ be the event that student will be pass in test-II. $X_{3}$ be the event that student will be pass in test-III. $\therefore P(X)=P\left(X_{1} \cap X_{2} \cap X_{3}^{\prime}\right)+P\left(X_{1} \cap X_{2}^{\prime} \cap X_{3}\right)$
$P\left(X_{1}^{1} \cap X_{2} \cap X_{3}\right).$
$\Rightarrow P(X)=P\left(X_{1}\right) \cdot P\left(X_{2}\right) \cdot P\left(X_{3}^{1}\right)+P\left(X_{1}\right) \cdot P\left(X_{2}^{\prime}\right) \cdot P\left(X_{3}\right)$
$+P\left(X_{1}^{1}\right) \cdot P\left(X_{2}\right) \cdot P\left(X_{3}\right)$
$\Rightarrow \quad \frac{1}{2}=p \cdot q \cdot \frac{1}{2}+p(1-q) \frac{1}{2}+p q \frac{1}{2}$
$\Rightarrow \quad \frac{1}{2}=p \cdot q \cdot \frac{1}{2}+p \cdot \frac{1}{2}-p \cdot q \cdot \frac{1}{2}+p \cdot q \cdot \frac{1}{2}$
$\Rightarrow \quad \frac{1}{2}=p \cdot \frac{1}{2}+p \cdot q \cdot \frac{1}{2}$
$\Rightarrow \quad(p+p q)=1$
$\Rightarrow \quad p(1+q)=1$
$P\left(X_{1}^{1} \cap X_{2} \cap X_{3}\right).$
$\Rightarrow P(X)=P\left(X_{1}\right) \cdot P\left(X_{2}\right) \cdot P\left(X_{3}^{1}\right)+P\left(X_{1}\right) \cdot P\left(X_{2}^{\prime}\right) \cdot P\left(X_{3}\right)$
$+P\left(X_{1}^{1}\right) \cdot P\left(X_{2}\right) \cdot P\left(X_{3}\right)$
$\Rightarrow \quad \frac{1}{2}=p \cdot q \cdot \frac{1}{2}+p(1-q) \frac{1}{2}+p q \frac{1}{2}$
$\Rightarrow \quad \frac{1}{2}=p \cdot q \cdot \frac{1}{2}+p \cdot \frac{1}{2}-p \cdot q \cdot \frac{1}{2}+p \cdot q \cdot \frac{1}{2}$
$\Rightarrow \quad \frac{1}{2}=p \cdot \frac{1}{2}+p \cdot q \cdot \frac{1}{2}$
$\Rightarrow \quad(p+p q)=1$
$\Rightarrow \quad p(1+q)=1$
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