Conditional probability problems and solutions pdf coins

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

You purchase a certain product. The manual states that the lifetime $T$ of the product, defined as the amount of time (in years) the product works properly until it breaks down, satisfies $$P(T \geq t)=e^>, \textrm < for all >t \geq 0.$$ For example, the probability that the product lasts more than (or equal to) $2$ years is $P(T \geq 2)=e^>=0.6703$. I purchase the product and use it for two years without any problems. What is the probability that it breaks down in the third year?

• Solution
  • Let $A$ be the event that a purchased product breaks down in the third year. Also, let $B$ be the event that a purchased product does not break down in the first two years. We are interested in $P(A|B)$. We have

  $P(B)$	$=P(T \geq 2)$
  $=e^>$.

  
  We also have
  

  $P(A)$	$=P(2 \leq T \leq 3)$
  $=P(T \geq 2)-P(T \geq 3)$
  $=e^>-e^>$.

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

  $P(A|B)$	$=\frac$
  $=\frac$
  $=\frac>-e^>>>>$
  $=0.1813$

  1. What is the probability of three heads, $HHH$?
  2. What is the probability that you observe exactly one heads?
  3. Given that you have observed at least one heads, what is the probability that you observe at least two heads?
  • Solution
    • We assume that the coin tosses are independent.
    • $P(HHH)=P(H)\cdot P(H) \cdot P(H)=0.5^3=\frac$.
    • To find the probability of exactly one heads, we can write

    $P(\textrm)$	$=P(HTT \cup THT \cup TTH)$
    $=P(HTT)+P(THT)+P(TTH)$
    $=\frac+\frac+\frac$
    $=\frac$.

    $P(A_2|A_1)$	$=\frac$
    $=\frac$
    $=\frac.\frac=\frac$.

    • $A$ and $C$ are independent,
    • $B$ and $C$ are independent,
    • $A$ and $B$ are disjoint,
    • $P(A \cup C)=\frac, P(B \cup C)=\frac, P(A \cup B\cup C)=\frac$

    

    • Solution
      • We can use the Venn diagram in Figure 1.26 to better visualize the events in this problem. We assume $P(A)=a, P(B)=b$, and $P(C)=c$. Note that the assumptions about independence and disjointness of sets are already included in the figure. Now we can write $$P(A \cup C)= a+c-ac=\frac;$$ $$P(B \cup C)=b+c-bc=\frac;$$ $$P(A \cup B\cup C)=a+b+c-ac-bc=\frac.$$ By subtracting the third equation from the sum of the first and second equations, we immediately obtain $c=\frac$, which then gives $a=\frac$ and $b=\frac$.
    • $A$ and $B$ are conditionally independent given $C_i$, for all $i \in \$;
    • $B$ is independent of all $C_i$'s.
    • Solution
      • Since the $C_i$'s form a partition of the sample space, we can apply the law of total probability for $A \cap B$:

      $P(A \cap B)$	$=\sum_^ P(A \cap B|C_i)P(C_i)$
      $=\sum_^ P(A|C_i)P(B|C_i)P(C_i) \hspace$	$\textrm< ($A$ and $B$ are conditionally independent)>$
      $=\sum_^ P(A|C_i)P(B)P(C_i)$	$\textrm< ($B$ is independent of all $C_i$'s)>$
      $=P(B) \sum_^ P(A|C_i)P(C_i)$
      $=P(B) P(A)$	$\textrm< (law of total probability).>$

      1. What is the probability that it's not raining and there is heavy traffic and I am not late?
      2. What is the probability that I am late?
      3. 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 event that there is heavy traffic, and $L$ be the event 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 to draw a tree diagram. Figure 1.27 shows a tree diagram for this problem. In this figure, each leaf in the tree corresponds to a single outcome in the sample space. We can calculate the probabilities of each outcome in the sample space by multiplying the probabilities on the edges of the tree that lead to the corresponding outcome.
        • The probability that it's not raining and there is heavy traffic and I am not late can be found using the tree diagram which is in fact applying the chain rule:

        $P(R^c \cap T \cap L^c)$	$= P(R^c)P(T|R^c)P(L^c|R^c \cap T)$
        $=\frac \cdot \frac \cdot \frac$
        $=\frac$.

        $P(L)$	$ = P(R,T,L)+P(R,T^c,L)+P(R^c,T,L)+P(R^c,T^c,L)$
        $=\frac+\frac+\frac+\frac$
        $= \frac$.

        $P(R \cap L)$	$ = P(R,T,L)+P(R,T^c,L)$
        $=\frac+\frac$
        $= \frac$.

        $P(R|L)$	$ = \frac$
        $=\frac. \frac$
        $= \frac$.

        • You pick a coin at random and toss it. What is the probability that it lands heads up?
        • You pick a coin at random and toss it, and 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 event that you choose a regular coin, and let $C_2$ be the event that you choose the two-headed coin. Note that $C_1$ and $C_2$ form a partition of the 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

          $P(H)$	$=P(H|C_1)P(C_1)+P(H|C_2)P(C_2)$
          $= \frac. \frac + 1 . \frac$
          $=\frac$.

          $P(C_2|H)$	$=\frac$
          $=\frac>>$
          $=\frac$.

          Here is another variation of the family-with-two-children problem [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 to find the 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 your result with the second part of Example 1.18.

          • Solution
            • Here we have four possibilities, $GG=(\textrm), GB, BG, BB$, and $P(GG)=P(GB)=P(BG)=P(BB)=\frac$. Let also $L$ be the event that the family has 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 find $P(GG|L)$:

            $P(GG|L)$	$=\frac$
            $= \frac$
            $= \frac<(2 \alpha-\alpha^2)\frac><(2 \alpha-\alpha^2)\frac+ \alpha \frac+ \alpha \frac+0.\frac>$
            $= \frac\approx \frac$.

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

            • Solution
              • Here again, we have four possibilities, $GG=(\textrm), GB, BG, BB$, and $P(GG)=P(GB)=P(BG)=P(BB)=\frac$. Now, let $G_r$ be the event that a randomly chosen child is a girl. Then we have $$P(G_r|GG)=1,$$ $$P(G_r|GB)=P(G_r|BG)=\frac,$$ $$P(G_r|BB)=0.$$ We can use Bayes' rule to find $P(GG|G_r)$:

              $P(GG|G_r)$	$=\frac$
              $= \frac$
              $= \frac>+ \frac \frac+ \frac \frac+0.\frac>$
              $= \frac$.

              Okay, another family-with-two-children problem. Just kidding! This problem has nothing to do with the two previous problems. I toss a coin repeatedly. The coin is unfair and $P(H)=p$. The game ends the first time that two consecutive heads ($HH$) or two consecutive tails ($TT$) are observed. I win if $HH$ is observed and lose if $TT$ is observed. For example if the outcome is $HTH\underline$, I lose. On the other hand, if the outcome is $THTHT\underline$, I win. Find the probability that I win.

              • Solution
                • Let $W$ be the event that I win. We can write down the set $W$ by listing all the different sequences that result in my winning. It is cleaner if we divide $W$ into two parts depending on the result of the first coin toss, $$W =\ \cup \.$$ Let $q=1-p$. Then

                $W$	$=P(\)+P(\)$
                $=p^2+p^3q+ p^4q^2+\cdots+p^2q+p^3q^2+ p^4q^3+\cdots$
                $=p^2(1+pq+(pq)^2+(pq)^3+\cdots)+p^2q(1+pq+(pq)^2+(pq)^3+\cdots)$
                $=p^2(1+q)(1+pq+(pq)^2+(pq)^3+\cdots)$
                $=\frac, \hspace\textrm< Using the geometric series formula>$
                $=\frac$.

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