# CÔNG SUẤT LÀ GÌ? CÔNG THỨC TÍNH CÔNG SUẤT VÀ BÀI TẬP VẬN DỤNG

In die & coin problems, unless stated otherwise, it is assumed coins & dice are fairand repeated trials are independent.

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You purchase a certain product. The manual states that the lifetime $T$ of the product,defined as the amount of time (in years) the sản phẩm works properly until it breaks down, satisfies$$P(T geq t)=e^-fract5, extrm for all t geq 0.$$For example, the probability that the hàng hóa lasts more than (or equal to) $2$ years is$P(T geq 2)=e^-frac25=0.6703$. I purchase the product và use it for two years withoutany problems. What is the probability that it breaks down in the third year?

Solution

Let $A$ be the sự kiện that a purchased hàng hóa breaks down in the third year. Also, let $B$ be the eventthat a purchased sản phẩm does not break down in the first two years. We are interested in $P(A|B)$. We have

$P(B)$ | $=P(T geq 2)$ |

$=e^-frac25$. |

**We also have**

Finally, since $A subset B$, we have $A cap B=A$. Therefore,

$P(A)$ | $=P(2 leq T leq 3)$ |

$=P(T geq 2)-P(T geq 3)$ | |

$=e^-frac25-e^-frac35$. |

$P(A|B)$ | $=fracP(A cap B)P(B)$ |

$=fracP(A)P(B)$ | |

$=frace^-frac25-e^-frac35e^-frac25$ | |

$=0.1813$ |

Problem

You toss a fair coin three times:What is the probability of three heads, $HHH$?What is the probability that you observe exactly one heads?Given that you have observed at least one heads, what is the probabilitythat you observe at least two heads?

Solution

We assume that the coin tosses are independent.$P(HHH)=P(H)cdot P(H) cdot P(H)=0.5^3=frac18$.To find the probability of exactly one heads, we can write

$P( extrmOne heads)$ | $=P(HTT cup THT cup TTH)$ |

$=P(HTT)+P(THT)+P(TTH)$ | |

$=frac18+frac18+frac18$ | |

$=frac38$. |

**Given that you have observed at least one heads, what is the probabilitythat you observe at least two heads? Let $A_1$ be the sự kiện that you observe atleast one heads, và $A_2$ be the sự kiện that you observe at least two heads. Then$$A_1 =S-TTT, extrm and P(A_1)=frac78;$$$$A_2 =HHT,HTH,THH,HHH, extrm và P(A_2)=frac48.$$Thus, we can write**

$P(A_2|A_1)$ | $=fracP(A_2 cap A_1)P(A_1)$ |

$=fracP(A_2)P(A_1)$ | |

$=frac48.frac87=frac47$. |

Problem

For three events $A$, $B$, & $C$, we know that$A$ & $C$ are independent,$B$ và $C$ are independent,$A$ & $B$ are disjoint,$P(A cup C)=frac23, P(B cup C)=frac34, P(A cup Bcup C)=frac1112$Find $P(A), P(B)$, and $P(C)$.

Solution

We can use the Venn diagram in Figure 1.26 khổng lồ bettervisualize the events in this problem. We assume $P(A)=a, P(B)=b$, & $P(C)=c$. Chú ý thatthe assumptions about independence and disjointness of sets are already included in the figure.

Fig.1.26 - Venn diagram for Problem 3.Now we can write$$P(A cup C)= a+c-ac=frac23;$$$$P(B cup C)=b+c-bc=frac34;$$$$P(A cup Bcup C)=a+b+c-ac-bc=frac1112.$$By subtracting the third equation from the sum of the first và second equations, weimmediately obtain $c=frac12$, which then gives $a=frac13$ & $b=frac12$.

**Problem**

Let $C_1, C_2,cdots,C_M$ be a partition of the sample space $S$, and $A$ & $B$ be two events.Suppose we know that$A$ & $B$ are conditionally independent given $C_i$, for all $i in 1,2,cdots,M$;$B$ is independent of all $C_i$"s.Prove that $A$ & $B$ are independent.

Solution

Since the $C_i$"s khung a partition of the sample space, we can apply the law oftotal probability for $A cap B$:

$P(A cap B)$ | $=sum_i=1^M P(A cap B|C_i)P(C_i)$ | |

$=sum_i=1^M P(A|C_i)P(B|C_i)P(C_i) hspace10pt$ | $ extrm ($A$ và $B$ are conditionally independent)$ | |

$=sum_i=1^M P(A|C_i)P(B)P(C_i)$ | $ extrm ($B$ is independent of all $C_i$"s)$ | |

$=P(B) sum_i=1^M P(A|C_i)P(C_i)$ | ||

$=P(B) P(A)$ | $ extrm (law of total probability).$ |

Problem

Problem

In my town, it"s rainy one third of the days. Given that it is rainy, there will be heavytraffic with probability $frac12$, và given that it is not rainy, there will be heavytraffic with probability $frac14$. If it"s rainy và there is heavy traffic, I arrivelate for work with probability $frac12$. On the other hand, the probability of beinglate is reduced khổng lồ $frac18$ if it is not rainy và there is no heavy traffic. Inother situations (rainy and no traffic, not rainy và traffic) the probability of being lateis $0.25$. You pick a random day.What is the probability that it"s not raining & there is heavy traffic & I am not late?What is the probability that I am late?Given that I arrived late at work, what is the probability that it rained that day?

Solution

Let $R$ be the event that it"s rainy, $T$ be the sự kiện that there is heavy traffic,and $L$ be the sự kiện that I am late for work. As it is seen from the problem statement,we are given conditional probabilities in a chain format. Thus, it is useful khổng lồ drawa tree diagram. Figure 1.27 shows a tree diagram for this problem.In this figure, each leaf in the tree corresponds to lớn a single outcome in the sample space.We can calculate the probabilities of each outcome in the sample space by multiplyingthe probabilities on the edges of the tree that lead to lớn the corresponding outcome.

Fig.1.27 - Tree diagram for Problem 5.The probability that it"s not raining và there is heavy traffic & I am not late canbe found using the tree diagram which is in fact applying the chain rule:

The probability that I am late can be found from the tree. All we need to vị is sumthe probabilities of the outcomes that correspond to lớn me being late. In fact, we are usingthe law of total probability here.

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We can find $P(R|L)$ using $P(R|L)=fracP(R cap L)P(L)$. We have already found$P(L)=frac1148$, & we can find $P(R cap L)$ similarly by adding the probabilitiesof the outcomes that belong lớn $R cap L$. In particular,

Thus, we obtain

Problem

$P(R^c cap T cap L^c)$ | $= P(R^c)P(T|R^c)P(L^c|R^c cap T)$ |

$=frac23 cdot frac14 cdot frac34$ | |

$=frac18$. |

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$P(L)$ | $ = P(R,T,L)+P(R,T^c,L)+P(R^c,T,L)+P(R^c,T^c,L)$ |

$=frac112+frac124+frac124+frac116$ | |

$= frac1148$. |

$P(R cap L)$ | $ = P(R,T,L)+P(R,T^c,L)$ |

$=frac112+frac124$ | |

$= frac18$. |

$P(R|L)$ | $ = fracP(R cap L)P(L)$ |

$=frac18. frac4811$ | |

$= frac611$. |

Problem

A box contains three coins: two regular coins & one giả two-headed coin ($P(H)=1$),You pick a coin at random và toss it. What is the probability that it lands heads up?You pick a coin at random and toss it, & get heads. What is the probability that it is the two-headed coin?

Solution

This is another typical problem for which the law of total probability is useful.Let $C_1$ be the sự kiện that you choose a regular coin, and let $C_2$ be the eventthat you choose the two-headed coin. Note that $C_1$ & $C_2$ size a partition ofthe sample space. We already know that$$P(H|C_1)=0.5,$$$$P(H|C_2)=1.$$Thus, we can use the law of total probability to write

$P(H)$ | $=P(H|C_1)P(C_1)+P(H|C_2)P(C_2)$ |

$= frac12. frac23 + 1 . frac13$ | |

$=frac23$. |

**Now, for the second part of the problem, we are interested in $P(C_2|H)$. We use Bayes" rule**

$P(C_2|H)$ | $=fracP(HP(H)$ |

$=frac1 . frac13frac23$ | |

$=frac12$. |

Problem

Here is another variation of the family-with-two-childrenproblem <1> <7>.A family has two children. We ask the father, "Do you have at least one daughter named Lilia?"He replies, "Yes!" What is the probability that both children are girls? In other words, we want khổng lồ findthe probability that both children are girls, given that the family has at least one daughter named Lilia.Here you can assume that if a child is a girl, her name will be Lilia with probability $alpha ll 1$independently from other children"s names. If the child is a boy, his name will not be Lilia. Compare yourresult with the second part of Example 1.18.

Solution

Here we have four possibilities, $GG=( extrmgirl, girl), GB, BG, BB$,and $P(GG)=P(GB)=P(BG)=P(BB)=frac14$. Let also $L$ be the sự kiện that the familyhas at least one child named Lilia. We have$$P(L|BB)=0,$$$$P(L|BG)=P(L|GB)=alpha,$$$$P(L|GG)=alpha (1-alpha)+ (1-alpha) alpha +alpha^2=2 alpha-alpha^2.$$We can use Bayes" rule to lớn find $P(GG|L)$:

$P(GG|L)$ | $=fracP(LP(L)$ |

$= fracP(LGG)P(GG)+P(L$ | |

$= frac(2 alpha-alpha^2)frac14(2 alpha-alpha^2)frac14+ alpha frac14+ alpha frac14+0.frac14$ | |

$= frac2-alpha4-alphaapprox frac12$. |

**Let"s compare the result with part (b) of Example 1.18. Amazingly, we notice that theextra information about the name of the child increases the conditional probability of$GG$ from $frac13$ khổng lồ about $frac12$. How can we explain this intuitively?Here is one way khổng lồ look at the problem. In part (b) of Example 1.18,we know that the family has at least one girl. Thus, the sample space reduces khổng lồ threeequally likely outcomes: $GG, GB, BG$, thus the conditional probability of $GG$ isone third in this case. On the other hand, in this problem, the available informationis that the sự kiện $L$ has occurred. The conditional sample space here still is $GG, GB,BG$, but these events are not equally likely anymore. A family with two girls is morelikely lớn name at least one of them Lilia than a family who has only one girl($P(L|BG)=P(L|GB)=alpha$, $ P(L|GG)=2 alpha-alpha^2$), thus in this case the conditionalprobability of $GG$ is higher. We would like to mention here that these problems are confusingand counterintuitive khổng lồ most people. So, vì not be disappointed if they seem confusing to lớn you.We seek several goals by including such problems.**

**First, we would like to emphasize that we should not rely too much on our intuition when solvingprobability problems. Intuition is useful, but at the end, we must use laws of probability khổng lồ solveproblems. Second, after obtaining counterintuitive results, you are encouraged khổng lồ think deeplyabout them lớn explain your confusion. This thinking process can be very helpful to lớn improve ourunderstanding of probability. Finally, I personally think these paradoxical-looking problems makeprobability more interesting.**

Problem If you are not yet confused, let"s look at another family-with-two-children problem! I know that afamily has two children. I see one of the children in the mall and notice that she is a girl.What is the probability that both children are girls? Again compare your result with the secondpart of Example 1.18.Note: Let"s agree on what precisely the problem statement means. Here is a more precise statement of theproblem: "A family has two children. We choose one of them at random và find out that she is a girl.What is the probability that both children are girls?"

Solution

Problem If you are not yet confused, let"s look at another family-with-two-children problem! I know that afamily has two children. I see one of the children in the mall and notice that she is a girl.What is the probability that both children are girls? Again compare your result with the secondpart of Example 1.18.Note: Let"s agree on what precisely the problem statement means. Here is a more precise statement of theproblem: "A family has two children. We choose one of them at random và find out that she is a girl.What is the probability that both children are girls?"

Solution

Here again, we have fourpossibilities, $GG=( extrmgirl, girl), GB, BG, BB$, and $P(GG)=P(GB)=P(BG)=P(BB)=frac14$.Now, let $G_r$ be the sự kiện that a randomly chosen child is a girl. Then we have$$P(G_r|GG)=1,$$$$P(G_r|GB)=P(G_r|BG)=frac12,$$$$P(G_r|BB)=0.$$We can use Bayes" rule to lớn find $P(GG|G_r)$:

$P(GG|G_r)$ | $=fracP(G_rP(G_r)$ |

$= fracP(G_rGB)P(GB)+P(G_r$ | |

$= frac1.frac141. frac14+ frac12 frac14+ frac12 frac14+0.frac14$ | |

$= frac12$. |

**So the answer again is different from the second part of Example 1.18. This is surprising tomost people. The two problem statements look very similar but the answers are completely different.This is again similar to lớn the previous problem (please read the explanation there). The conditionalsample space here still is $GG, GB, BG$, but the point here is that these are not equally likelyas in Example 1.18. The probability that a randomly chosen childfrom a family with two girls is a girl is one, while this probability for a family who hasonly one girl is $frac12$. Thus, intuitively, the conditional probability of the outcome$GG$ in this case is higher than $GB$ and $BG$, and thus this conditional probability must belarger than one third.**

Problem

Problem

Okay, another family-with-two-children problem. Just kidding! This problem has nothing to do with thetwo previous problems. I toss a coin repeatedly. The coin is unfair & $P(H)=p$. The trò chơi ends thefirst time that two consecutive heads ($HH$) or two consecutive tails ($TT$) are observed. I win if$HH$ is observed & lose if $TT$ is observed. For example if the outcome is $HTHunderlineTT$, Ilose. On the other hand, if the outcome is $THTHTunderlineHH$, I win. Find the probability that Iwin.

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SolutionLet $W$ be the event that I win. We can write down the mix $W$ by listing all the differentsequences that result in my winning. It is cleaner if we divide $W$ into two parts dependingon the result of the first coin toss,$$W =HH, HTHH, HTHTHH,cdots cup THH, THTHH, THTHTHH,cdots .$$Let $q=1-p$. Then

$W$ | $=P(HH, HTHH, HTHTHH,cdots )+P(THH, THTHH, THTHTHH,cdots )$ |

$=p^2+p^3q+ p^4q^2+cdots+p^2q+p^3q^2+ p^4q^3+cdots$ | |

$=p^2(1+pq+(pq)^2+(pq)^3+cdots)+p^2q(1+pq+(pq)^2+(pq)^3+cdots)$ | |

$=p^2(1+q)(1+pq+(pq)^2+(pq)^3+cdots)$ | |

$=fracp^2(1+q)1-pq, hspace10pt extrm Using the geometric series formula$ | |

$=fracp^2(2-p)1-p+p^2$. |