���urm�9�aj��D���]���ց����V�q/�C2B����X��c 1��R~��}6�K��5?>��� $P(HHH)=P(H)\cdot P(H) \cdot P(H)=0.5^3=\frac{1}{8}$. All we need to do is sum immediately obtain $c=\frac{1}{2}$, which then gives $a=\frac{1}{3}$ and $b=\frac{1}{2}$. The manual states that the lifetime $T$ of the product, Here is another variation of the family-with-two-children as in Example 1.18. \frac{2}{3} + 1 . defined as the amount of time (in years) the product works properly until it breaks down, satisfies from a family with two girls is a girl is one, while this probability for a family who has understanding of probability. What is the probability that it is the two-headed coin? Example (Genetics) Traits passed from generation to generation are carried by genes. $$W =\{HH, HTHH, HTHTHH,\cdots \} \cup \{THH, THTHH, THTHTHH,\cdots \}.$$ First, we would like to emphasize that we should not rely too much on our intuition when solving extra information about the name of the child increases the conditional probability of $$P(H|C_1)=0.5,$$ You pick a random day. about them to explain your confusion. Here we have four possibilities, $GG=(\textrm{girl, girl}), GB, BG, BB$, If the child is a boy, his name will not be Lilia. and $P(GG)=P(GB)=P(BG)=P(BB)=\frac{1}{4}$. Let $R$ be the event that it's rainy, $T$ be the event that there is heavy traffic, $=\frac{e^{-\frac{2}{5}}-e^{-\frac{3}{5}}}{e^{-\frac{2}{5}}}$, $=\sum_{i=1}^{M} P(A|C_i)P(B|C_i)P(C_i) \hspace{10pt}$, $\textrm{ ($A$ and $B$ are conditionally independent)}$, $\textrm{ ($B$ is independent of all $C_i$'s)}$, $=\frac{2}{3} \cdot \frac{1}{4} \cdot \frac{3}{4}$, $ = P(R,T,L)+P(R,T^c,L)+P(R^c,T,L)+P(R^c,T^c,L)$, $=\frac{1}{12}+\frac{1}{24}+\frac{1}{24}+\frac{1}{16}$, $= \frac{1}{2}. A family with two girls is more We have already found 0 This ... Marilyn gave a solution concluding that you should switch, and if you do, your probability of winning is 2/3. has at least one child named Lilia. $P(L)=\frac{11}{48}$, and we can find $P(R \cap L)$ similarly by adding the probabilities BG$, but these events are not equally likely anymore. A box contains three coins: two regular coins and one fake two-headed coin ($P(H)=1$), This is another typical problem for which the law of total probability is useful. $$P(G_r|GB)=P(G_r|BG)=\frac{1}{2},$$ visualize the events in this problem. result with the second part of Example 1.18. lose. $= \frac{P(G_r|GG)P(GG)}{P(G_r|GG)P(GG)+P(G_r|GB)P(GB)+P(G_r|BG)P(BG)+P(G_r|BB)P(BB)}$, $= \frac{1.\frac{1}{4}}{1. Problem . What is the probability that both children are girls? is that the event $L$ has occurred. Let $q=1-p$. The aim of this chapter is to revise the basic rules of probability. We assume that the coin tosses are independent. the law of total probability here. Here again, we have four h��Y�NI����|�ֈ�}�ZHؘ�����ș�\Y�M}�%�Ȣ�E�0܌���5ND)�b!�R�JJK�2 UXO-�P2Q��J{�,TN#�_S�/��D�B'E|�0N�Ra��J&E��K)]�`p5e The two problem statements look very similar but the answers are completely different. on the result of the first coin toss, We assume $P(A)=a, P(B)=b$, and $P(C)=c$. The probability that it's not raining and there is heavy traffic and I am not late can f{�Hi_ �?h#�"؅�������9@� ��� Suppose we know that. problem [1] [7]. the probabilities of the outcomes that correspond to me being late. Read Free Conditional Probability Practice Problems With Solutions Probability Probability Conditional Probability 19 / 33 Conditional Probability Example Example De ne events B 1 and B 2 to mean that Bucket 1 or 2 was selected and let events R, W, and B indicate if the color of the ball is red, white, or black. Given that I arrived late at work, what is the probability that it rained that day? only one girl is $\frac{1}{2}$. Each gene is of one of the types C(dominant gene) or c(recessive gene). Okay, another family-with-two-children problem. So, do not be disappointed if they seem confusing to you. This thinking process can be very helpful to improve our we are given conditional probabilities in a chain format. %PDF-1.6 %���� possibilities, $GG=(\textrm{girl, girl}), GB, BG, BB$, and $P(GG)=P(GB)=P(BG)=P(BB)=\frac{1}{4}$. I win if probability of $GG$ is higher. Figure 1.27 shows a tree diagram for this problem. Here you can assume that if a child is a girl, her name will be Lilia with probability $\alpha \ll 1$ win. We already know that As it is seen from the problem statement, Second, after obtaining counterintuitive results, you are encouraged to think deeply Let $C_1, C_2,\cdots,C_M$ be a partition of the sample space $S$, and $A$ and $B$ be two events. we know that the family has at least one girl. endstream endobj 1171 0 obj <>/Metadata 74 0 R/Names 1199 0 R/Outlines 196 0 R/PageMode/UseNone/Pages 1166 0 R/StructTreeRoot 233 0 R/Type/Catalog>> endobj 1172 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/StructParents 0/Tabs/S/Type/Page>> endobj 1173 0 obj <>stream In Then we have total probability for $A \cap B$: In my town, it's rainy one third of the days. equally likely outcomes: $GG, GB, BG$, thus the conditional probability of $GG$ is I toss a coin repeatedly. We can use the Venn diagram in Figure 1.26 to better On the other hand, if the outcome is $THTHT\underline{HH}$, I win. 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conditional probability problems with solutions pdf

\frac{1}{4}+ \frac{1}{2} \frac{1}{4}+ \frac{1}{2} \frac{1}{4}+0.\frac{1}{4}}$, $=P(\{HH, HTHH, HTHTHH,\cdots \})+P(\{THH, THTHH, THTHTHH,\cdots \})$, $=p^2+p^3q+ p^4q^2+\cdots+p^2q+p^3q^2+ p^4q^3+\cdots$, $=p^2(1+pq+(pq)^2+(pq)^3+\cdots)+p^2q(1+pq+(pq)^2+(pq)^3+\cdots)$, $=\frac{p^2(1+q)}{1-pq}, \hspace{10pt}\textrm{ Using the geometric series formula}$. 7�bᵥ�TxknReqZ��jZ>jܸ6,��~�{�����(�����W�/f��jv������F���8��V�O�5( Zŗ��|(fA���\�_����re��QӉ��i�>/g��'+��/�����ϯ�˓r��x&���x%�%ވC�V�!�ʼn8��X��sq>�N��B����Kq ՏR\�o��J\�JL�T��\�g�X�E������|Z�Z��r&��X����,��Ϲ�?ğ⧸�����W��v��`���U]XRdzg�_w�)v�s*�Fp��T���#Ea��xZ��w��~����g���V�Iu�7�^)NV����8�]�4�^V��|)>5ڲ���~�Ώ���QI&���(��5� .���:ڄ48�H�F���+�"�S0��s��fT͊��t�e��f�u�S73f;# ��p��,kHi[9a4�F�"Ȁ|�BK ��(a�F��Fi�T1Dk�)�H�D:��h�5L�%�D:�l�$�>���urm�9�aj��D���]���ց����V�q/�C2B����X��c 1��R~��}6�K��5?>��� $P(HHH)=P(H)\cdot P(H) \cdot P(H)=0.5^3=\frac{1}{8}$. All we need to do is sum immediately obtain $c=\frac{1}{2}$, which then gives $a=\frac{1}{3}$ and $b=\frac{1}{2}$. The manual states that the lifetime $T$ of the product, Here is another variation of the family-with-two-children as in Example 1.18. \frac{2}{3} + 1 . defined as the amount of time (in years) the product works properly until it breaks down, satisfies from a family with two girls is a girl is one, while this probability for a family who has understanding of probability. What is the probability that it is the two-headed coin? Example (Genetics) Traits passed from generation to generation are carried by genes. $$W =\{HH, HTHH, HTHTHH,\cdots \} \cup \{THH, THTHH, THTHTHH,\cdots \}.$$ First, we would like to emphasize that we should not rely too much on our intuition when solving extra information about the name of the child increases the conditional probability of $$P(H|C_1)=0.5,$$ You pick a random day. about them to explain your confusion. Here we have four possibilities, $GG=(\textrm{girl, girl}), GB, BG, BB$, If the child is a boy, his name will not be Lilia. and $P(GG)=P(GB)=P(BG)=P(BB)=\frac{1}{4}$. Let $R$ be the event that it's rainy, $T$ be the event that there is heavy traffic, $=\frac{e^{-\frac{2}{5}}-e^{-\frac{3}{5}}}{e^{-\frac{2}{5}}}$, $=\sum_{i=1}^{M} P(A|C_i)P(B|C_i)P(C_i) \hspace{10pt}$, $\textrm{ ($A$ and $B$ are conditionally independent)}$, $\textrm{ ($B$ is independent of all $C_i$'s)}$, $=\frac{2}{3} \cdot \frac{1}{4} \cdot \frac{3}{4}$, $ = P(R,T,L)+P(R,T^c,L)+P(R^c,T,L)+P(R^c,T^c,L)$, $=\frac{1}{12}+\frac{1}{24}+\frac{1}{24}+\frac{1}{16}$, $= \frac{1}{2}. A family with two girls is more We have already found 0 This ... Marilyn gave a solution concluding that you should switch, and if you do, your probability of winning is 2/3. has at least one child named Lilia. $P(L)=\frac{11}{48}$, and we can find $P(R \cap L)$ similarly by adding the probabilities BG$, but these events are not equally likely anymore. A box contains three coins: two regular coins and one fake two-headed coin ($P(H)=1$), This is another typical problem for which the law of total probability is useful. $$P(G_r|GB)=P(G_r|BG)=\frac{1}{2},$$ visualize the events in this problem. result with the second part of Example 1.18. lose. $= \frac{P(G_r|GG)P(GG)}{P(G_r|GG)P(GG)+P(G_r|GB)P(GB)+P(G_r|BG)P(BG)+P(G_r|BB)P(BB)}$, $= \frac{1.\frac{1}{4}}{1. Problem . What is the probability that both children are girls? is that the event $L$ has occurred. Let $q=1-p$. The aim of this chapter is to revise the basic rules of probability. We assume that the coin tosses are independent. the law of total probability here. Here again, we have four h��Y�NI����|�ֈ�}�ZHؘ�����ș�\Y�M}�%�Ȣ�E�0܌���5ND)�b!�R�JJK�2 UXO-�P2Q��J{�,TN#�_S�/��D�B'E|�0N�Ra��J&E��K)]�`p5e The two problem statements look very similar but the answers are completely different. on the result of the first coin toss, We assume $P(A)=a, P(B)=b$, and $P(C)=c$. The probability that it's not raining and there is heavy traffic and I am not late can f{�Hi_ �?h#�"؅�������9@� ��� Suppose we know that. problem [1] [7]. the probabilities of the outcomes that correspond to me being late. Read Free Conditional Probability Practice Problems With Solutions Probability Probability Conditional Probability 19 / 33 Conditional Probability Example Example De ne events B 1 and B 2 to mean that Bucket 1 or 2 was selected and let events R, W, and B indicate if the color of the ball is red, white, or black. Given that I arrived late at work, what is the probability that it rained that day? only one girl is $\frac{1}{2}$. Each gene is of one of the types C(dominant gene) or c(recessive gene). Okay, another family-with-two-children problem. So, do not be disappointed if they seem confusing to you. This thinking process can be very helpful to improve our we are given conditional probabilities in a chain format. %PDF-1.6 %���� possibilities, $GG=(\textrm{girl, girl}), GB, BG, BB$, and $P(GG)=P(GB)=P(BG)=P(BB)=\frac{1}{4}$. I win if probability of $GG$ is higher. Figure 1.27 shows a tree diagram for this problem. Here you can assume that if a child is a girl, her name will be Lilia with probability $\alpha \ll 1$ win. We already know that As it is seen from the problem statement, Second, after obtaining counterintuitive results, you are encouraged to think deeply Let $C_1, C_2,\cdots,C_M$ be a partition of the sample space $S$, and $A$ and $B$ be two events. we know that the family has at least one girl. endstream endobj 1171 0 obj <>/Metadata 74 0 R/Names 1199 0 R/Outlines 196 0 R/PageMode/UseNone/Pages 1166 0 R/StructTreeRoot 233 0 R/Type/Catalog>> endobj 1172 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/StructParents 0/Tabs/S/Type/Page>> endobj 1173 0 obj <>stream In Then we have total probability for $A \cap B$: In my town, it's rainy one third of the days. equally likely outcomes: $GG, GB, BG$, thus the conditional probability of $GG$ is I toss a coin repeatedly. We can use the Venn diagram in Figure 1.26 to better On the other hand, if the outcome is $THTHT\underline{HH}$, I win.

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