Saturday, May 31, 2008

Probability: More Practice

Instead of starting a new unit, we worked on Probability. Mostly we went over the practice sheet we were given on Thursday. The questions we reviewed were one, two, six and nine.

Question One:

  • The probability of a randomly chosen car being defective is 1/3. Four cars are chosen randomly in order. Given that at least two cars are defective, what is the probability that the first car is defective?

The first thing we did, was figure out what the question was asking.
P(1st is defective I
at least two is defective)

Next we found the sample space. It's always easier if you think of it as a word, and then asking yourself how many times the letters in that word can be rearranged, like so:
DDGG -- meaning two of the cars are defective
DDDG -- three cars are defective
DDDD -- all four cars are defective
Then you can find the probability that the cars will be chosen that way, and that becomes your sample space, or denominator.

Next we looked at all the possibilities that the first car will be defective using slots, which, if you remember, was from combinatorics.

Put that numerator on top of the denominator that we found earlier, and you have your answer.

Question Two:

  • Two jars contain red and green marbles. Jar I contains 3 red and 2 green marbles. Jar II contains 4 red and 3 green marbles. A jar is picked at random and two marbles are picked out of that jar in order. If it is known that the first marble is red, what is the probability of the second marble being red?
You can start this problem by drawing a tree, and writing down the probability that each possibility will occur - if that makes sense. For example, if you look at the picture above, you'll see that the first branch is 'Jar I' followed by 'Jar II', the probability of choosing either jar is one half. The next branches are 'red' and 'green', and the probability of choosing red in 'Jar I' is 3/5, while the probability of choosing green is 2/5. So on and so forth. Keep in mind that the marbles are not replaced after you take it out, so the sample space is different everytime you draw.

In this question, what we're looking for is the probability of getting a red after drawing a red marble first. That becomes the numerator, and the denominator consists of all the possibilities, as shown above. So you plug in the proper numbers in their proper places, and you have that question solved.

Question Six:

  • Susan sees her friend, Tim, at his locker with a worried look on his face. She asks, "What's wrong?" Tim has to open his locker and change clothes within the next five minutes. However, he has forgotten the combination to his new lock. He knows that the lock requires three different numbers. He also remembers that all of the numbers are odd, and all of the numbers are divisible by seven. It takes 10 seconds to dial a locker combination and 1.5 minutes to change clothes. Is Tim likely to be ready for gym class on time?
I think you'll understand how we solved this problem just by looking at the above picture. The '1 min 30 sec', written in pink, is how long it takes Tim to change.

Question Nine:
  • Three identical boxes each contain two drawers. In one box, each drawer contains a gold coin. In another box, each drawer contains a silver coin. The remaining box has a silver coin in one drawer and a gold coin in the other. One drawer is opened and a gold coin is found. What is the probability that the other drawer in that box also contains a gold coin.
  • Michael claims that the probability is 1/3. Jessica claims it is 1/2. Raymond says the probability is 2/3. Explain how each person may have arrived at their answer. Who is correct? Justify your answer.

The probability of choosing the right box is 1/3, and the probability of choosing either a gold or silver coin is 1/2, since there's two drawers. Obviously if you get the box that only have two gold coins, you're not going to get a silver coin. But I'm sure we're all smart enough to realize that.

The probability of getting a second gold coin is shown by:
[P(IG)/(P(IIIG) + P(IG) + P(IG))] + [P(IG)/(P(IIIG) + P(IG) + P(IG))]

If we plug in the correct numbers, we have:
[(1/6)/((1/6) + (1/6) + (1/6))] + [(1/6)/((1/6) + (1/6) + (1/6))]
[(1/6)/(1/2)] + [(1/6)/(1/2)]
(1/3) + (1/3)

Therefore, Raymond is correct. Michael could've gotten his answer by thinking that there's only one box that contains two gold coins, and the probability of choosing that box is 1/3. Jessica could've simply thought, well there's two drawers, so there's a 1/2 chance of getting a gold coin.

So that's the end of that. There's also questions in the slides (9-18) that we could do for practice. In the afternoon class we had a pre-test that we're going to go over tomorrow. There isn't going to be a test for this unit because we don't have time. Soo.. we're starting a new unit next week.

And that's that. The next scribe will be Joseph because he asked me yesterday. Whoot. Good luck to everyone doing the English Provincial Exams tomorrow! Remember to pace yourself and.. uhh, I forgot what else Mr. V and Mrs. Strecker said.. but.. yes!


Rakiztah said...
This comment has been removed by the author.
m@rk said...


Great scribe post! I am really impress on how thoroughly explained your scribe post is. I really struggled in probability ever since grade 9 and is one of the reasons why i took this class again. Its too bad to hear that we are not going to have a test for this unit, because i really want to measure my expertise in this. BTW, good luck on your provincial exam tomorrow.