Probability of success given another success (conditional probability problem)

Question

Suppose that independent students have to solve a homework. Student A solves the homework with a probability of $70\%$, Student B solves it with a probability of $80\%$ and Student C solves it with a probability of $90\%$.

a) What is the probability that all the students solve the homework if at least two students solve the homework?

b) What is the probability that Student B solves the homework if less than two students solve the homework?

c) What is the probability that less than two students solve the homework?

I have tried this procedure:

a) There are three possible cases:

• Student B and C have already solved the homework, and so student A has yet to solve the problem ($70\%$)
• Student A and C have already solved the homework, and so student B has yet to solve the problem ($80\%$)
• Student A and B have already solved the homework, and so student C has yet to solve the problem ($90\%$)

The probability of the first event is ($80\% * 90\% = 72\%$), for the second is ($70\% * 90\% = 63\%$), for the third is ($70\% * 80\% = 56\%$). So the result is: ($70\% * 72\%$) + ($80\% * 63\%$) + ($90\% * 56\%$) but the result is unrealistic ($151.2\%$). What is wrong?

b) I had in mind to use a procedure very similar to point a) but I believe that I get an wrong result

c) The overall probability is given by: ($70\% * 20\% * 10\%$) + ($30\% * 80\% * 10\%$) + ($30\% * 20\% * 90\%$) = $9.2\%$ Is it correct?

Answer 1

I'll assume that by team $A$, you meant student $A$, and similarly for $B,C$.

Define events $A,B,C,X,Y$ as

• $A$ is the event that student $A$ solves the homework.$\\[4pt]$
• $B$ is the event that student $B$ solves the homework.$\\[4pt]$
• $C$ is the event that student $C$ solves the homework.$\\[4pt]$
• $X$ is the event that all three students solve the homework.$\\[4pt]$
• $Y$ is the event that at least two students solve the homework.

and for any event $E$, let $E'$ denote the complementary event "$\text{not}\,E$".

Then we have $$ P(X)=P(A)P(B)P(C)=\frac{7}{10}\cdot \frac{4}{5}\cdot \frac{9}{10}=\frac{63}{125}=.504 $$ and \begin{align*} P(Y)&= P(A\cap B\cap C') \,+\, P(B\cap C\cap A') \,+\, P(C\cap A\cap B') \,+\, P(A\cap B\cap C) \\[4pt] &= \left(\frac{7}{10}\cdot \frac{4}{5}\cdot \frac{1}{10}\right) + \left(\frac{4}{5}\cdot \frac{9}{10}\cdot \frac{3}{10}\right) + \left(\frac{9}{10}\cdot \frac{7}{10}\cdot \frac{1}{5}\right) + \left(\frac{7}{10}\cdot \frac{4}{5}\cdot \frac{9}{10}\right) \\[4pt] &= \frac{451}{500} = .902 \\[4pt] \end{align*} For parts $(a)$ and $(b)$, you need to apply the formula for conditional probability.

Part $(a)$:$\;$The probability that all three students solve the homework given that at least two students solve the homework is $$ P(X|Y) = \frac{P(X\cap Y)}{P(Y)} = \frac{P(X)}{P(Y)} = \frac {\left({\large{\frac{63}{125}}}\right)} {\left({\large{\frac{451}{500}}}\right)} = \frac{252}{451} \approx .5587583149 $$ Part $(b)$:$\;$The probability that student $B$ solves the homework given that less than two students solve the homework is $$ P(B|Y') = \frac{P(B\cap Y')}{P(Y')} = \frac{P(B\cap C'\cap A')}{P(Y')} = \frac { \left( {\large{\frac{4}{5}}} \cdot {\large{\frac{1}{10}}} \cdot {\large{\frac{3}{10}}} \right) } {\left({\large{\frac{49}{500}}}\right)} = \frac{12}{49} \approx .2448979592 $$ For part $(c)$, you were on the right track but you missed the case where none of the three students solve the homework.

But since we've already computed $P(Y)$, there's an easier way . . .

Part $(c)$:$\;$The probability that less than two students solve the homework is $$ P(Y') = 1-P(Y) = 1-\frac{451}{500} = \frac{49}{500} = .098 $$