Question

A and B are two students .Their chances of solving a problem correctly are 1/3 and 1/4.If the probability of their making a common error is 1/20 and they obtain the same answer then the probability of their answer to be correct is?

A.1/12 B.1/40 C.13/120 D.10/13

Answer 1

Task. A and B are two students. Their chances of solving a problem correctly are 1/3 and 1/4. If the probability of their making a common error is 1/20 and they obtain the same answer then the probability of their answer to be correct is?

A. 1/12

B. 1/40

C. 13/120

D. 10/13

Solution. Consider the following events:

$C_{A}=``$A solved problem correctly”

$C_{B}=``$B solved problem correctly”

S = “A and B obtain the same answer”

C = “Their answer is correct”

E = “A and B made common error”.

We will assume that the events $C_{A}$, $C_{B}$, and $E$ independent, that is

$P(C_{A}C_{B}E)=P(C_{A})*P(C_{B})*P(E),$

$P(C_{A}C_{B})=P(C_{A})*P(C_{B}),\qquad P(C_{A}E)=P(C_{A})*P(E),\qquad P(C_{B}E)=P(C_{B})*P(E).$

We should find the conditional probability $P(C|S)$. By definition where SC is the intersection of events S and C:

SC = “A and B obtain correct answer”

By assumption we have that

$P(C_{A})=\frac{1}{3},\qquad P(C_{B})=\frac{1}{4},\qquad P(E)=\frac{1}{20}.$

We will express the probabilities $P(SC)$ and $P(S)$ via $P(C_{A})$, $P(C_{B})$, and $P(\overline{C_{A}}\,\overline{C_{B}})$.

Notice that events SC and $C_{A}C_{B}$ coincide:

SC = $C_{A}C_{B}=``$A and B obtain correct answer”,

so

$P(SC)=P(C_{A}C_{B})=P(C_{A})*P(C_{B})=\frac{1}{3}*\frac{1}{4}=\frac{1}{12}.$

Now compute $P(S)$. Notice that $S$ happens if either of the following two events hold:

$C_{A}C_{B}=``A$ and $B$ made no errors’

$\overline{C_{A}C_{B}}E=``A$ and $B$ made error and this error is the same”.

These events are mutually exclusive, that is

$P(S)=P(C_{A}C_{B})+P(F).$

Moreover,

$P(F)=P(\overline{C_{A}C_{B}}E).$

Since events $C_{A}$, $C_{B}$ and $E$ are independent, we obtain that

$P(F)$ $=P(\overline{C_{A}C_{B}}E)==P(\overline{C_{A}})*P(\overline{C_{B}})*P(E)$

$=\left(1-\frac{1}{3}\right)*\left(1-\frac{1}{4}\right)*\frac{1}{20}$ $=\frac{2}{3}*\frac{3}{4}*\frac{1}{20}=\frac{1}{40},$

so

$P(S)=P(C_{A}C_{B})+P(F)=\frac{1}{3}*\frac{1}{4}+\frac{1}{40}=\frac{1}{12}+\frac{1}{40}=\frac{13}{120}.$

Therefore

$P(C|S)=\frac{P(SC)}{P(S)}=\frac{P(C_{A}C_{B})}{P(C_{A}C_{B})+P(F)}=\frac{\frac{1}{12}}{\frac{13}{120}}=\frac{10}{13}.$

Answer. D) $\frac{10}{13}$.