A student which knows nothing, went to a test includes 6 questions with answers 'yes' or 'no' only. Find the probability that:

a.The student got 100

b.The student got 100 if he knew there are 3 question with answer yes and 3 with answer no.

c.The student got 100 if he knew (I) there are 3 question with answer yes and 3 with answer no (II) doesn't exist sequence of 3 question with same answer (e.g answers for 3,4,5 are no and for 1,2,6 yes).

Let's define the following events: $A-\text{student got 100},B-\text{student knows there are 3 yes asnwers and 3 no's},C-\text{Doesn't exist sequence of 3 same answers}$

about a: It seems clear that probability for a correct answer is $\frac 1 2$,and the answers are independent 'bernouli tests' so the probability is $\frac 1 {64}$.

About B: I think I need to use definition of conditioned probability ($P(A\mid B)=\frac{P(A\cap B)}{P(B)}$) but I don't $P(B)$. I assume that $P(A\cap B)$ is binomial, means choosing 3 of 6 questions and multiplying by $(\frac 1 2)^3\cdot (\frac 1 2)^3$ getting $P(A\cap B)=\frac 5 {16}$.

About C: I think its almost the same but here I don't know both $P(C)$ and $P((A\cap B)\cap C)$.

I'll be glad for help finding $P(B),P(C),P(A\cap B\cap C)$.


You have the right mindset here.

1). Yes. This is the correct application of $6$ bernoulli tests, aka, Binomial$(6,.5)$ Using the binomial formula provides the answer you have received because $$P(X=6)={6 \choose 6}(.5)^6(.5)^0=\frac{1}{64}.$$

2). Let event $A$ be the event of getting 100. Let event $B$ be the event of knowing there are $3$ Y's and N's. Thus, $A\cap{B}$ is the event you get 100 and you know there are $3$ Y's and $3$ N's. So this probability $P(A\cap{B})$ is $\frac{1}{64}$, because, thinking intuitively, the only possibility in the intersection of these two events is the event of getting $100$. Anything less ($2$ Y's and $4$ N's for example) is not $100$ From application of binomial with 3 successes out of 6, we get that $P(B)={6 \choose 3}(.5)^3(.5)^3=20\cdot\frac{1}{64}=\frac{20}{64}$. Using the definition of conditional probability then we can see that $$P(A|B)=\frac{P(A\cap{B})}{P(B)}=\frac{\frac{1}{64}}{\frac{20}{64}}=\frac{1}{20}$$

3). You can use this same idea for part 3, though I think the best way to find the members of event $C$ is observation. So let's look at all $20$ possible permuations of (YYYNNN). $$ \begin{matrix} YYYNNN & YYNYNN & YYNNYN & YYNNNY & YNYNNY\\ YNNYNY & YNNNYY & YNYYNN & YNNYYN & YNYNYN\\ NNNYYY & NNYNYY & NNYYNY & NNYYYN & NYNYYN\\ NYYNYN & NYYYNN & NYNNYY & NYYNNY & NYNYNY\\ \end{matrix} $$ So it is clear from just simply looking out our $20$ permuations that we have $14$ different ways of not getting $3$ Y's or N's in a row. So essentially this is $P(B\cap{C})$. $P(A\cap{B}\cap{C})$ is exactly the same as $P(A\cap{B})$ because the intersection of all these events can $only$ be event $A$, so $P(A\cap{B}\cap{C})=\frac{1}{64}$. And $P(B\cap{C})$ is the event of knowing there are $3$ Y's and $3$ N's $and$ no string of three of the same answers is possible is $\frac{14}{64}$ since $C\subset{B}$ So again this is also conditional. Thus $$P(A|B\cap{C})=\frac{P(A\cap{B}\cap{C})}{P(B\cap{C})}=\frac{\frac{1}{64}}{\frac{14}{64}}=\frac{1}{14}$$

Intuitively this makes complete sense, since in each new scenario, we are finding out we know more about the exam and leaving less to chance. Thus as we know more, probability of getting 100 increases.

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  • $\begingroup$ You wrote YNNNYY and probably another one twice. Then, you probably also miss 2 possibilities. $\endgroup$ – N. S. Sep 16 '13 at 14:15
  • $\begingroup$ Thank you! I will amend...i missed the YNYNYN (NYNYNY) case...that makes 14. $\endgroup$ – Eleven-Eleven Sep 16 '13 at 14:23

B There are $\binom{6}{3}=20$ ways of distributing 3 yes on 6 positions. Thus there are 20 possible answer sheets with 3 yes and 3 no.

C There are $20$ possible answers containing 3 yes and 3 no. There are 4 of them which contain 3 consecutive yes (the yes can start at possitions...) and 4 which contain three consecutive no, but we counted the YYYNNN and NNNYYY twice. Thus there are 20-8+2=14 possible answer in this case and only one is good.

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