Question

Let $P_{1},P_{2}$ and $P_{3}$ are the probabilities of a student passing three independent exams $A,B$ and $C$ respectively. If $P_{1},P_{2}$ and $P_{3}$ are the roots of equation $20x^{3}-27x^{2}+14x-2=0$ , then the probability that the student passes in exactly one of $A,B$ and $C$ is

• A. $\frac{3}{20}$
• B. $\frac{7}{20}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (C)

$P_{1}+P_{2}+P_{3}=\frac{27}{20}$
$P_{1}P_{2}+P_{2}P_{3}+P_{3}P_{1}=\frac{14}{20}$
$P_{1}P_{2}P_{3}=\frac{2}{20}$
Probability that the student passes in exactly one of $A,B,C$ is
$=P_{1}+P_{2}+P_{3}-2\left(P_{1} P_{2} + P_{2} P_{3} + P_{3} P_{1}\right)+3P_{1}P_{2}P_{3}$
$=\frac{27}{20}-2\left(\frac{14}{20}\right)+\frac{6}{20}=\frac{5}{20}=\frac{1}{4}$