Question : A speaks the truth 5 out of 7 times and B speaks the truth 8 out of 9 times. What is the probability that they contradict each other in stating the same fact?
Option 1: $\frac{1}{7}$
Option 2: $\frac{1}{9}$
Option 3: $\frac{1}{4}$
Option 4: $\frac{1}{3}$
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Correct Answer: $\frac{1}{3}$
Solution : Probability of A speaking truth = $\frac{5}{7}$ = P(A) ⇒ Probability of A lying = $1-\frac{5}{7}$ = $\frac{2}{7}$ = P(A') Probability of B speaking truth = $\frac{8}{9}$ = P(B) ⇒ Probability of B lying = $1-\frac{8}{9}$ = $\frac{1}{9}$ = P(B') $\therefore$ Required probability = P(A) × P(B′) + P(A′) × P(B) = $\frac{5}{7}×\frac{1}{9}+\frac{2}{7}×\frac{8}{9}$ = $\frac{5}{63}+\frac{16}{63}$ = $\frac{21}{63}$ = $\frac{1}{3}$ Hence, the correct answer is $\frac{1}{3}$.
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Question : The value of $3 \frac{1}{5} \div 4 \frac{1}{2}$ of $5 \frac{1}{3}+\frac{1}{8} \div \frac{1}{2}$ of $\frac{1}{4}-\frac{1}{4}\left(\frac{1}{2} \div \frac{1}{8} \times \frac{1}{4}\right)$ is:
Question : A person can hit a target 5 times out of 8 shots. If he fires 10 shots, what is the probability that he will hit the target twice?
Question : Find the fraction which bears the same ratio to $\frac{1}{27}$ that $\frac{3}{7}$ does to $\frac{5}{9}$.
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