Date: 07/26/98 at 10:12:22 From: charla anderson Subject: Probabilities A machine makes bottle caps and packs them automatically into boxes. One box in ten has a defective bottle cap. a) What is the probability that each of the next four boxes will have a defective bottle cap? b) What is the probability that two boxes out of the first four will have a defective bottle cap? Thanks for your help, Charla
Date: 07/26/98 at 17:28:32 From: Doctor Ken Subject: Re: Probabilities Hi Charla, These kinds of probability problems can get pretty tricky, but if you master them, they can be pretty satisfying. One way to approach these problems is to use probability trees. Here's how they work: Let's say the probability of having a sunny day is 0.7, i.e. there's a 70 percent chance that each day will be sunny (this is a funny weather system in which one day's weather has no effect on the next day's weather). I can draw a tree like this: start / \ / \ 0.7/ \0.3 / \ sunny cloudy /\ /\ / \ / \ 0.7/ \0.3 0.7/ \0.3 / \ / \ sunny cloudy sunny cloudy Now, what does this diagram mean? Well, we can use it as a tool to find probabilities. For instance, what's the probability of having two sunny days in a row? We start at the top, choose the sunny path, and choose the sunny path again: start // \ // \ 0.7// \0.3 // \ sunny cloudy //\ /\ // \ / \ 0.7// \0.3 0.7/ \0.3 // \ / \ sunny cloudy sunny cloudy Now we multiply all the numbers we passed along the way: 0.7 * 0.7 = 0.49. That means there's a 49% chance of ending up at the sunny-sunny "leaf" on the tree. If you want to, you can do all the leaves' multiplications in advance: start / \ / \ 0.7/ \0.3 / \ sunny cloudy /\ /\ / \ / \ 0.7/ \0.3 0.7/ \0.3 / \ / \ sunny cloudy sunny cloudy 0.49 0.21 0.21 0.09 Now, what's the probability of having a sunny day and then a cloudy day? Reading the number at the end of the sunny-cloudy path tells us that it's 0.21. Piece of cake! So what's the probability of having at least one sunny day in two days? Well, we have to figure out which "leaves" on our tree satisfy the condition. Sunny-sunny, sunny-cloudy, and cloudy-sunny do, but cloudy- cloudy doesn't. So we add up the probabilities of the three leaves: 0.49 + 0.21 + 0.21 = 0.91, which is the answer. Now let's see if we can tackle your problem. The probability of any box having a defective bottle cap in it is 0.1 (1 divided by 10). So we can make a tree. Since your problems ask about sequences of four boxes, we'll make it four branches long. I'll use g for good and b for bad: start / \ / \ 0.9/ \0.1 / \ g b / \ / \ 0.9/ \0.1 0.9/ \0.1 / \ / \ g b g b / \ / \ / \ / \ 0.9/ \0.1 0.9/ \0.1 0.9/ \0.1 0.9/ \0.1 / \ / \ / \ / \ g b g b g b g b .9/\0.1 0.9/\0.1 0.9/\0.1 0.9/\0.1 0.9/\0.1 0.9/\0.1 0.9/\0.1 0.9/\.1 / \ / \ / \ / \ / \ / \ / \ / \ g b g b g b g b g b g b g b g b You can use the same method as in my example: to find the probability that each of the next four boxes will have a defective cap, follow the b-b-b-b branch. To find the probability that two of the next four boxes will have bad caps, look at all the branches with exactly two g's and two b's. I'll let you figure out which branches those are (there are six of them). Then add those branches' final numbers together. Good luck! - Doctor Ken, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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