Exercise books for all videos: http://shop.strandmathe.de/ If you carry out several random experiments one after the other, you get a multi-stage random experiment. Models like this can be clearly presented using tree diagrams. Draw three times from the urn shown without putting it back. Calculate the probabilities in the random experiment above for P( , , ), P( , , ) and P( , , ). The events A = "Fewer than two red balls were drawn" and B = "At least one blue ball was drawn" are also given. Use the path rules to determine the corresponding probabilities for the events. The 2nd path rule is needed for the probabilities of events. You should first note down the result paths that fulfill the event. For event A: (b,b,b); (b,b,r); (b,r,b); (r,b,b) and for event B: (b,r,r); (r,b,r); (r,r,b); (b,b,r); (b,r,b); (b,r,b); (b,b,b); Let's start with P(A). Using the 2nd path rule, the probability of event A occurring can be calculated in two steps. First, we need the probabilities of the outcomes: P(b,b,b)=5/42 ; P(b,b,r)=10/63 and because the 'r' in the other two outcomes was simply 'permuted' to a different place, 10/63 is also the probability of the third and fourth outcome paths, i.e. P(b,r,b)=10/63 and P(r,b,b)=10/63 . In the second step, these individual probabilities must be added together. P(A)=5/42+10/63+10/63+10/63=25/42≈0.5952=59.52% . For event B we can either proceed as we did for event A, or use a trick. We know that the probability of event B added to the probability of the opposite event B ̅ is exactly '1' (i.e. P(B)+P(B ̅ )=1). The only path that does not satisfy event B is (r,r,r). This means that P(B ̅ )=P(r,r,r)=1/21 . If we rearrange the above formula, we get P(B)=1-P(B ̅ )=1-P(r,r,r). We therefore get P(B)=1-1/21=20/21≈0.9524=95.24% . Trainer: "When drawing the tree diagram, the branches must always add up to '1' and the denominator of the probabilities must always be '1' smaller from one level to the next if you draw without replacing." 1. Forrest Gump is the main character in the Hollywood film of the same name. His mother always said: Life is like a box of chocolates - you never know what you're going to get. Suppose we have a box of 12 chocolates. Three of them contain alcohol (a), four contain nuts (n) and five contain pistachio (p). For you and your friend, you reach into the box of chocolates twice at random and without replacing. Create a two-level tree diagram for this random experiment. Calculate the probability of first drawing a chocolate with nuts and then a chocolate with pistachio (n,p). Calculate the probability of first drawing a chocolate with pistachio and then a chocolate with nuts (p,n). What do you notice? Determine the probabilities P(E1) and P(E2) with E1 = "No chocolates with alcohol" and E2 = "At least one pistachio". 2. On the basis of past statistical data, a meteorologist can determine that there is a 3/4 probability that a rainy day in October will be followed by another rainy day. If it is rain-free on one day in October, there is a 2/5 probability that the following day will also be rain-free. Today is Wednesday, October 10th, and it is raining. What is the probability of having a rain-free day while hiking next Saturday? State whether the probability is greater or less than 10% that it will rain at most once by Sunday. Today's rainy day is not one of them. 3. In penalty shootouts, Müller scores 85%, Götze 70%, and Hector 64% of the time. In the World Cup they will play against Italy in this order. What is the probability that at least one of them will score. Tip: Calculate with the opposite probability or the opposite event. Facebook: / strandmathe Instagram: / strandmathe Twitter: / strandmathe