Student $A_1$ will solve a problem with probability $0.85$, and student $A_2$ with probability $0.9$.

$1)$ Find the probability that a problem will be solved if students solve it independently.

$2)$ Find the probability that both students will solve the problem if they solve it independently.

$3)$ Students were solving a problem separately. If a problem is solved by one of the students, find the probability that student $A_1$ solved it.


We define three events:

$H_1:$ Student $A_1$ will solve a problem.

$H_2:$ Student $A_2$ will solve a problem.

$A:$ Problem is solved.

We are given that $P(H_1)=0.85,P(H_2)=0.9$.

$1)$ If we use the formula of total probability, for event $A$: $$P(A)=P(A|H_1)P(H_1)+P(A|H_2)P(H_2).$$

Conditional probabilities $P(A|H_1),P(A|H_2)$ are not given.

Is it necessary to find these conditional probabilities?

Using the equation for union of two events: $$P(A)=P(H_1\cup H_2)=P(H_1)+P(H_2)-P(H_1)P(H_2)=0.985$$

$2)$ $P(H_1\cap H_2)=P(H_1)P(H_2)=0.765$

$3)$ By using Bayes theorem formula:

$$P(H_1|A)=\frac{P(A\cap H_1)}{P(A)}=\frac{P(A|H_1)P(H_1)}{P(A|H_1)P(H_1)+P(A|H_2)P(H_2)}$$

In $1)$ we found that $P(A)=P(H_1\cup H_2)=P(H_1)+P(H_2)-P(H_1)P(H_2)=0.985$.

Is it a mistake?

If we set $P(A|H_1)=P(A)$, then $P(H_1|A)=P(H_1)=0.85$.

Are $1),2),3)$ correctly solved?

  • 1
    $\begingroup$ Note: it is highly unrealistic to expect that the two events, student $A_1$ , $A_2$ get it right, are independent. Obviously if the problem is trivial it is more likely that they'll both get it and if the problem is very hard it is more likely that neither will. $\endgroup$ – lulu Jan 10 '17 at 18:38

Problems 1 and 2 are right, but 3 is not.

For 3, you are asked for the probability that student 1 solves the problem given that it was solved by either student. In your notation this is $$P(H_1\mid H_1\cup H_2) = \frac{P(H_1\cap(H_1\cup H_2))}{P(H_1\cup H_2)}.$$ By drawing a picture or reasoning otherwise, you can see that $H_1\cap (H_1\cup H_2) = H_1,$ so the answer is $$\frac{P(H_1)}{P(H_1\cup H_2)} = \frac{0.85}{0.985} = 0.863.$$

Make sure this answer makes sense. Does it make sense that it should be larger than 0.85? Is is surprising that it isn't much larger? If instead of .9, student 2 had an 100% probability of solving the problem, it would be 0.85. Does that make sense?

EDIT: Commenter below points out that the question is probably better interpreted as "if the problem is solved by exactly one of the students, what is the probability it is solved by student $A_1$." This is very different. For that, you want $$P(H_1|(H_1\cap H_2^c)\cup (H_1^c\cap H_2))= \frac{P(H_1\cap((H_1\cap H_2^c)\cup (H_1^c\cap H_2)))}{P((H_1\cap H_2^c)\cup (H_1^c\cap H_2))}$$

Similarly to before, the top gives $P(H_1-H_1\cap H_2) = .85-.765 = 0.085.$ The bottom is the probability that only one of them solves it, so is $P(H_1)+P(H_2)-2P(H_1\cap H_2) = 0.22.$ The answer is then $0.085/0.22 = 0.368$

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    $\begingroup$ For the third question, isn't it an exclusive disjunction ? $\endgroup$ – Canardini Jan 10 '17 at 18:32
  • $\begingroup$ @Canardini You mean like is the question "if the problem is solved, what is the probabillity that student 1 solved it but not student 2?" That's not how I read it. $\endgroup$ – spaceisdarkgreen Jan 10 '17 at 18:35
  • $\begingroup$ The other way, if problem is solved by "exactly" one student, what is the probability that $A_1$ solved it. I may be wrong. $H_1 \cup H_2$ includes the event that both students solved it $\endgroup$ – Canardini Jan 10 '17 at 18:39
  • $\begingroup$ @Canardini Could you show the equation for exclusive disjunction probability? $\endgroup$ – user300045 Jan 10 '17 at 18:51
  • $\begingroup$ $$P(H_1 | \{H_1 \cap H_2^c\} \cup \{H_1^c \cap H_2\})$$ $\endgroup$ – Canardini Jan 10 '17 at 18:56

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