Question

Three candidates solve a question. Odds in favour of the correct answer are $5:2,4:3$ and $3:4$ respectively for the three candidates. What is the probability that at least two of them solve the question correctly?

• A. $\frac{209}{343}$
• B. $\frac{134}{343}$
• C. $\frac{149}{343}$
• D. $\frac{60}{343}$

Answer 1

The correct option is A $\frac{209}{343}$

Let A,B,C be the three students.

Let P(A), P(B), P(C) be the probabilities that students A,B and C answer the question correctly.

$P\left(A\right)=\frac{5}{5+2}=\frac{5}{7}$

$P\left(B\right)=\frac{4}{4+3}=\frac{4}{7}$

$P\left(C\right)=\frac{3}{3+4}=\frac{3}{7}$

Probability that atleast 2 of them solve the question correctly

$=P(A\cap B\cap \overline{C})+P(A\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap C)+P(A\cap B\cap C)$

$=P\left(A\right)P\left(B\right)P\left(\overline{C}\right)+P\left(A\right)P\left(\overline{B}\right)P\left(C\right)+P\left(\overline{A}\right)P\left(B\right)P\left(C\right)+P\left(A\right)P\left(B\right)P\left(C\right)$

$=\left(\frac{5}{7}\right)\left(\frac{4}{7}\right)(1-\frac{3}{7})+\left(\frac{5}{7}\right)(1-\frac{4}{7})\left(\frac{3}{7}\right)+(1-\frac{5}{7})\left(\frac{4}{7}\right)\left(\frac{3}{7}\right)+\left(\frac{5}{7}\right)\left(\frac{4}{7}\right)\left(\frac{3}{7}\right)$

$=\left(\frac{5}{7}\right)\left(\frac{4}{7}\right)\left(\frac{4}{7}\right)+\left(\frac{5}{7}\right)\left(\frac{3}{7}\right)\left(\frac{3}{7}\right)+\left(\frac{2}{7}\right)\left(\frac{4}{7}\right)\left(\frac{3}{7}\right)+\left(\frac{5}{7}\right)\left(\frac{4}{7}\right)\left(\frac{3}{7}\right)$

$=\frac{(5\times 4\times 4)+(5\times 3\times 3)+(2\times 4\times 3)+(5\times 4\times 3)}{7\times 7\times 7}$

$=\frac{209}{243}$

Answer is option (A)