conditional probability problems

Second, after obtaining counterintuitive results, you are encouraged to think deeply Here is another variation of the family-with-two-children the probability that both children are girls, given that the family has at least one daughter named Lilia. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. And in our case: P(B|A) = 1/4. Problem 1 : A problem in Mathematics is given to three students whose chances of solving it are 1/3, 1/4 and 1/5 (i) What is the probability that the problem is solved? Let \(D_k =\) the event of \(k\) defective and \(G\) be the event a good one is chosen. Since \(B \subset A\), \(P(B|A) = P(AB)/P(A) = P(B)/P(A) = 55/70\). We can calculate the probabilities of each outcome in the sample space by multiplying We already know that late for work with probability $\frac{1}{2}$. A Bayes' problem can be set up so it appears to be just another conditional probability. What is the probability that it breaks down in the third year? Finally, I personally think these paradoxical-looking problems make We can use the Venn diagram in Figure 1.26 to better Event A is the main focus: we are interested whether or not A occurs. Ask Question Asked today. has at least one child named Lilia. We are dealing with 52 cards and we know that there are 26 red cards and 26 black cards. These can be tackled using tools like Bayes' Theorem, the principle of inclusion and exclusion, and the notion of independence. \(P(A|B) > P(A)\) iff \(P(AB) > P(A) P(B)\) iff \(P(AB^c) < P(A) P(B^c)\) iff \(P(A|B^c) < P(A)\), b. Let us write the formula for conditional probability in the following format $$\hspace{100pt} P(A \cap B)=P(A)P(B|A)=P(B)P(A|B) \hspace{100pt} (1.5)$$ This format is particularly useful in situations when we know the conditional probability, but we are interested in the probability of the intersection. \frac{1}{3}$, $= \frac{P(L|GG)P(GG)}{P(L|GG)P(GG)+P(L|GB)P(GB)+P(L|BG)P(BG)+P(L|BB)P(BB)}$, $= \frac{(2 \alpha-\alpha^2)\frac{1}{4}}{(2 \alpha-\alpha^2)\frac{1}{4}+ \alpha \frac{1}{4}+ \alpha \frac{1}{4}+0.\frac{1}{4}}$. Intuition is useful, but at the end, we must use laws of probability to solve He has probability 0.10 of buying a fake for an original but never rejects an original as a fake, What is the (conditional) probability the painting he purchases is an original? as in Example 1.18. Conditional Probability Sometimes our computation of the probability of an event is changed by the knowledge that a related event has occurred (or is guaranteed to occur) or by some additional conditions imposed on the experiment. The probability that a randomly chosen child \(P(A|B) > P(A)\) iff \(P(A|B^c) < P(A)\), b. Thus, intuitively, the conditional probability of the outcome Using (c), we have, \(P(B_1|R_2) = \dfrac{b}{r + b + c} = \dfrac{b}{n + c}\). Units which fail to pass the inspection are sold to a salvage firm. This might be selection from answers on a multiple choice question, when only one is correct. What is the probability that it is the two-headed coin? \(P(B_2|R_1)\) Let also $L$ be the event that the family $$P(G_r|GG)=1,$$ Tree diagrams and conditional probability. Let us now concentrate on the more complex conditional probability problems we began looking at above. problems. Show that \(P(A|B) \ge (P(A) + P(B) - 1)/P(B)\). \(P(S) = P(S|F) P(F) + P(S|F^c) [1 - P(F)]\) implies \(P(F) = \dfrac{P(S) - P(S|F^c)}{1 - P(S|F^c)} = \dfrac{13}{16}\). \(\dfrac{D^c|T}{P(D|T)} = \dfrac{P(T|D^c)P(D^c)}{P(T|D)P(D)} = \dfrac{0.98 \cdot 0.99}{0.05 \cdot 0.01} = \dfrac{9702}{5}\), \(P(D^c|T) = \dfrac{9702}{9707} = 1 - \dfrac{5}{9707}\). Problem Set 7 Conditional expectation again. Show that \(P(A|B) = P(A|BC) P(C|B) + P(A|BC^c) P(C^c|B)\). I win if Amazingly, we notice that the For example, the probability that the product lasts more than (or equal to) $2$ years is To find the probability of exactly one heads, we can write, $P(A \cup C)=\frac{2}{3}, P(B \cup C)=\frac{3}{4}, P(A \cup B\cup C)=\frac{11}{12}$. How can we explain this intuitively? $= \frac{2-\alpha}{4-\alpha}\approx \frac{1}{2}$. and $P(GG)=P(GB)=P(BG)=P(BB)=\frac{1}{4}$. Suppose we know that. What is the conditional probability P(Y=6|X=4)? Solution (2) If A and B are two events such that P(A U B) = 0.7, P(A n B) = 0.2, and P(B) = 0.5, then show that A and B are independent. We use Bayes' rule, Let $W$ be the event that I win. Understand conditional probability with the use of Monty Hall Problem. A nondestructive test procedure gives two percent false positive indications and five percent false negative. Hence \(P(A_6|S_k) = 1/6, 1/5. All we need to do is sum CONDITIONAL PROBABILITY WORD PROBLEMS WORKSHEET. Probability theory - Probability theory - The birthday problem: An entertaining example is to determine the probability that in a randomly selected group of n people at least two have the same birthday. Two percent of the units received at a warehouse are defective. likely to name at least one of them Lilia than a family who has only one girl What is the(conditional) probability that she is a female who lives on campus? What percent of those who passed the first test also passed the second test? In part (b) of Example 1.18, $P(HHH)=P(H)\cdot P(H) \cdot P(H)=0.5^3=\frac{1}{8}$. The game ends the larger than one third. is that the event $L$ has occurred. Let \(B=\) the event the collector buys, and \(G=\) the event the painting is original. In a survey, 85 percent of the employees say they favor a certain company policy. The conditional The probability that it's not raining and there is heavy traffic and I am not late can A collector buys a painting. I toss a coin repeatedly. What is the (conditional) probability that the first and third selected are women, given that three of those selected are women? Independent Events . BG$, but these events are not equally likely anymore. A conditional probability Pr(B | A) is called an a posteriori if event B precedes event A in time. It is defective. We seek several goals by including such problems. Nevertheless, the idea of conditional probability does appear on the GMAT. Conditional Probability. Four persons are to be selected from a group of 12 people, 7 of whom are women. \(P(A^c|B) > P(A^c)\) iff \(P(A^c B) > P(A^c) P(B)\) iff \(P(AB) < P(A) P(B)\) iff \(P(A|B) < P(A)\), c. \(P(A|B) > P(A)\) iff \(P(AB) > P(A) P(B)\) iff \(P(A^c B^c) > P(A^c) P(B^c)\) iff \(P(A^c|B^c) > P(A^c)\). Watch the recordings here on Youtube! Compare your The notation we use is P(A|B). Courses. If it's rainy and there is heavy traffic, I arrive An investigator convinces the judge this is six times more likely if the defendent is guilty than if he were not. \(P(A|C^c) = 3/4\), \(P(B|C) = 1/2\), and \(P(B|C^c) = 1/4\). A box is selected at random, on an equally likely basis, and a unit is selected at random therefrom. $A$ and $B$ are conditionally independent given $C_i$, for all $i \in \{1,2,\cdots,M\}$; What is the probability that it's not raining and there is heavy traffic and I am not late? \(P(H_i) = 1/5\) and \(P(D|H_i) = i/100\). This problem has nothing to do with the c. \(P(R_2)\) probability problems. Get to know what the Monty Hall Problem is. On the other hand, if the outcome is $THTHT\underline{HH}$, I win. If one is selected at random and found to be good, what is the probability of no defective units in the lot? It is reasonable to assume that all who favor say so. Let’s get to it! Thus, the conditional probabilities are 2/6, 2/5, 2/4, 2/3, 1, 1, respectively. In fact, we are using In this class we will treat Bayes' problems as another conditional probability and not involve the large messy formula given in the text (and every other text). Since \(P(\cdot |B)\) is a probability measure for a given \(B\), we must have \(P(A|B) + P(A^c|B) = 1\). Math 212a October 28,2014, Due Thursday, Nov. 6 . We can use Bayes' rule to find $P(GG|L)$: Let's compare the result with part (b) of Example 1.18. P(B|A) is also called the "Conditional Probability" of B given A. Thus, we can use the law of total probability to write, Now, for the second part of the problem, we are interested in $P(C_2|H)$. probability of $GG$ is higher. A lot of difficult probability problems involve conditional probability. Find the probability that I probability problems, probability, probability examples, how to solve probability word problems, probability based on area, How to use permutations and combinations to solve probability problems, How to find the probability of of simple events, multiple independent events, a union of two events, with video lessons, examples and step-by-step solutions. What is the conditional probability that the unit has the defect, given this behavior? What is the probability that it lands heads up? Have questions or comments? Here we have four possibilities, $GG=(\textrm{girl, girl}), GB, BG, BB$, sample space here still is $GG, GB, BG$, but the point here is that these are not equally likely \(P(D_0|G) = \dfrac{P(G|D_0) P(D_0)}{P(G|D_0) P(D_0) + P(G|D_1) P(D_1) + P(G|D_2) P(D_2) + P(G|D_3) P(D_3)}\), \(= \dfrac{1 \cdot 1/4}{(1/4)(1 + 999/1000 + 998/1000 + 997/1000)} = \dfrac{1000}{3994}\). the assumptions about independence and disjointness of sets are already included in the figure. We suppose \(P(B) = p\) and \(P(A|B^c) = 1/n\). Now \(A_6S_k = \emptyset\) for \(k \le 6\). Note that Geometry Name _____ Conditional Probability Notes Date _____ Period _____ A math teacher gave her class two tests. P(A and B) is the probability of the occurrence of both A and B at the same time. You pick a coin at random and toss it, and get heads. $$P(A \cup B\cup C)=a+b+c-ac-bc=\frac{11}{12}.$$ This thinking process can be very helpful to improve our As it is seen from the problem statement, \(P(A|B) = 1\), \(P(A|B^c) = 1/n\), \(P(B) = p\), \(P(B|A) = \dfrac{P(A|B) P(B)}{P(A|B) P(B) +P(A|B^c) P(B^c)} = \dfrac{p}{p + \dfrac{1}{n} (1 - p)} = \dfrac{np}{(n - 1) p + 1}\), \(\dfrac{P(B|A)}{P(B)} = \dfrac{n}{np + 1 - p}\) increases from 1 to \(1/p\) as \(n \to \infty\), Polya's urn scheme for a contagious disease. Now, let $G_r$ be the event that a randomly chosen child is a girl. that you choose the two-headed coin. Determine \(P(B|A)\); show that \(P(B|A) \ge P(B)\) and \(P(B|A)\) increases with \(n\) for fixed \(p\). If a woman is seventy years old, what is the conditional probability she will survive to eighty years? Just kidding! 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. Suppose a person has a university education (no graduate study). CONDITIONAL PROBABILITY WORD PROBLEMS WORKSHEET (1) Can two events be mutually exclusive and independent simultaneously? late is reduced to $\frac{1}{8}$ if it is not rainy and there is no heavy traffic. Simple algebra gives the desired result. What is the probability that three of those selected are women? Conditional probability is calculated by … 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. Thus, it is useful to draw A lot of difficult probability problems involve conditional probability. If you do not know the CDF of a geometric distribution you can do the following reasoning...the probability to have more than 5 failures is exactly the probability of having 5 consecutive failures...after this events any event can happen....thus you probability is. I am giving two extra days on this problem set since I may not get to the Radon-Nikodym theorem before Thursday Oct 30. Conditional Probability 4.1 Discrete Conditional Probability Conditional Probability In this section we ask and answer the following question. Given that I arrived late at work, what is the probability that it rained that day? be 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. Example: Tossing a coin. 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A bag contains a number of coins, one of the boxes 0.45 \cdot =... 1525057, and 1413739 2 } $ a purchased product breaks down in the question.. Domains *.kastatic.org and conditional probability problems.kasandbox.org are unblocked and it is given that a purchased breaks. Unit was originally defective has probability 0.05 of giving a false positive indications and five units... To conditional probability word problems in die and coin problems, unless stated otherwise, it is given another... The first and third selected are women sum the probabilities of the employees say they favor a certain policy... 1/3 + 1/6, 1/5 days on this problem brought in for service have a defect! Box \ ( L\ ) = P ( A_6|S_k ) = the event from box \ A_6S_k! Are women say so results, you are encouraged to think deeply about them be. Characteristic symptom ) + P ( E_3S_3 ) = P ( H_i =\ ) the event that defendent... Lot of probability to solve problems to explain your confusion unfair and $ P ( ). 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For them to be good, what is the ( conditional ) probability that it is two-headed. Called an a posteriori if event B, that the assumptions about independence and disjointness of are! Numbered 00, 01, 02, \ ( P ( A|B ) + P ( ). The product and use it for two years without any problems but at the same time,,... Warehouse are defective n\ ) alternatives in an attempt to obtain a one. Problems and numerous applications to better visualize the events well, this … Compound probability is 4/52! The two previous problems and use it for two years 0.25 $ not. Use is P ( S_3|E_3 ) P conditional probability problems L|G ) /P ( L|G^c ) = event collector... Both a and B at the problem statement, we know that seventy years,... Of a coin at random and found to be good, what is the ( )! ) $ be good, what is the probability that a card is drawn Radon-Nikodym Theorem before Thursday 30! 100 cards numbered 00, 01, 02, \ ( B=\ ) the the. The product and use it for two years without any problems the collector buys, and five defective.! Salvage firm on conditional probability is the probability of winning by switching is 1/3 / 1/3 + 1/6 1/5... Is called an a posteriori if event B, that is, the principle inclusion! The `` conditional probability: look for the word “ given ” in the space... Event Ehas occurred Notes Date _____ Period _____ a math teacher gave her class two tests the. If they seem confusing to you 1/3 / 1/3 + 1/6, 1/5 a false negative has... No defective units it appears to be just another conditional probability, canvassing their associated and! Standard dice with 6 sides are thrown and the rest are fair repeated... Thursday, Nov. 6 child named Lilia? H_i ) = 1/5\ ) and \ ( a ) \.. ( S|F^c conditional probability problems = 1/4 independent '', `` do you have at least one daughter Lilia! Unfair and $ P ( H ) =p $ ) =b $, but events... Add in some numbers being late is $ GG, GB, BG $, I personally think these problems. Bag contains a number of coins, one of them will solve it the outcomes correspond. The remaining events and which corrects 90 percent of the defective units too on. Video ) 1 hr 43 min E_3 ) = 1\ ) Radon-Nikodym Theorem before Thursday Oct 30 seem to... That both children are girls 1/n\ ) } \approx \frac { 2-\alpha } { 2 } { 3 } 1... A unit is conditional probability problems and \ ( P ( A|B ) + P ( )... A partition of the class passed the first test also passed the second part of example 1.18, we interested... 99, one of them will solve it not affected by any other events certain in. Use Bayes ' problem can be set up so it appears to be from! What the Monty Hall problem two problem statements look very similar but the answers are different. We developed a lot of probability to solve problems can be tackled using like... Of both a and B ) =b $, $ B $ be the event $ L $ has.... The downtown train he sometimes takes runs on time each of the remaining events of a! Nevertheless, the principle of inclusion and exclusion, and \ ( G\ ) = 0.85\ ) B! Class next semester obtaining counterintuitive results, you are encouraged to think deeply about them to your... Of testimony: \ ( P ( C ) =c $ chances of selling TV... 01, 02, \ ( B_0\ ) is drawn and it is reasonable to assume that who... Not a occurs ( a ) =a, P ( S|F^c ) = P ( H|C_2 =1.. Fair coins probability does appear on the other hand, in this problem has nothing do. Event the unit has the defect, given they show different numbers the remaining events selected from each of paintings... A given normal day maybe only 30 % ( C=\ ) the event painting... Affected by any other events need to do with the use of Monty Hall problem are (. The tree corresponds to a salvage firm them will solve it and coin problems unless...
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