Scribe post : the end of probability
well first all I would like to "thank" Abriel for picking me. wow my class is on a roll here..7 hall of famers in a row. CONGRATS! Now all the pressure is on me loL:( better make this good.
today was a two period day and was also the day we wrapped up the probability unit. In the first period class we went straight to our groups from yesterday and finished working on the problems.
#2) From a group of 8 men and 7 women, a committee of 6 is chosen. What is the probability that 3 men and 3 women will be chosen?
7C38C3 / 15C6
- 7C3 is the number of possible ways women can be chosen. The 7 represents the number of women and the 3 represents how many can be chosen.
- 8C3 is the number of possible ways men can be chosen. The number 8 represents the number of men and the 3 represents how many can be chosen.
- 15C6 is the number of possible ways with no restrictions. This is the sample space.
#3) Frankie take 1 language course, 1 math couse and 1 science course. The probability of Frankie passing a language course is 0.9, a math course is 0.7 and a science course is 0.6 frankies father picks one course at random and asks "Frankie did you pass?"
A) What is the probability that the correct answer is "Yes"?
- M= math
- F= French
In this question you are trying to figure out the probability of Frankie saying yes so you multiply the probabilities for the subjects and passing.Once you have those answers you add them together and you get you answer for the question.
P(MP) + P(FP) + P(SP)
= 7/30 + 4/30 + 6/30
B) If the correct answer is "Yes", what is the probability that the course was Math?
P(MP) / P(MP) + P(FP) + P(SP)
= (7/30) / (22/30)
- P(MP) + P(FP) + P(SP) is the sample space which we got from the previous question. Therefore this is the denominator. This is the sample space because this is all the possible ways Frankie can answer "yes" .
- P(MP) is what you want to figure out so it is the numerator
#2) Eight students of different heights are seated randomly around a circular table. Find the probability that the two tallest students are sitting next to each other.
step1) There are 8 students all together and they are going to sit in a circle. Because they are going to sit in a circle you have to set a reference point.
(8-1)! = 7!
step2) The question also states that the two tallest boy are going to sit together so you have to put them in a bag and count them as one. Because of this you are left with only 6 bags.
(7-1)! = 6! (you get the 7 from the previous step)
step 3) Now you have to figure out the number of ways the two people in the bag can sit.
step 4) now you figure out the sample space because this is a probability question. This is the number of all possible outcomes.
step 5) put it all together.
6!2! / 8!
After finishing the group work from yesterday we started on some other questions while still working in out groups.
Suppose a test for cancer is known to be 98% accurate (ie. The outcome of the test is correct 98% of the time). Suppose 0.5% of the population has cancer. What is the probability that a person who tests positive had cancer?
Suppose 1,000,000 randomly selected people are tested. 4 things can happen.
- a person with cancer tests positive
- a person with cancer test negative
- a heathly person test positive
- a healthy person test negative
#1a) How many of the tested people have cancer?
(number of people)(percent of people with cancer) = people with cancer
(1,000,000)(0.005) = 5000
b) How many do not have cancer?
(number of people) - (people with cancer) = people with no cancer
(1,000,000) - (5000) = 995,000
(percent)( number of people) = people with no cancer
(0.995)(1,000,000) = 995, 000
#2) the test is 98% accurate
a) how many people with cancer will test positive?
(people with cancer)(0.98) = accurate count
(5000)(0.98) = 4900
b) how many people with no cancer will test negative?
(people with cancer) - (people who will test positive) = people with no cancer who will test negative
(5000) - (4900) = 100
(percent that's not accurate)(people with cancer) = people with no cancer who will test negative
(0.02)(5000) = 100
#3a) How many healthy people will test positive?
(number of healthy people)(percent that's not accurate) = healthy people who will test positive
(995,000)(0.02) = 19,900
b) How many healthy people will test negative?
(number of healthy people)-(healthy people who will test positive) = healthy people will test negative
#4a) How many people tested positive for cancer?
#2a + #3a = people who tested positive for cancer
4900 + 19900 = 24800
b) How many of those people have cancer?
c) What is the probability that a person who tests positive for cancer had cancer?
P(C(given that)P) = 4900 / ( 4900 + 19900 ) = 19.76
To get these answers you just simply multiply along the branches
P(C(given that)P) = P(CP) / (P(CP) + P(HP)
=0.0049 / (0.0049+0.0199)
The afternoon started with some more probability questions...
A company makes computer chips, 70% of all chips come from factory A and 30% from factory B. In factory A 25% of the chips are defective. In factory B 10% are defective.
a) Suppose we do not know what factory a chip came from. What is the probability it is defective?
P(D) = P(AD) + P(BD)
=0.175 + 0.03
In this question you are trying to look for the probability of the chips being defective. So in order to do this you add the defective probabilities for factory A and Factory B
B)You found a defective ship! What is the probability it came from factory A?
P(A(given that)D)= P(AD) / P(D)
=0.175 / 0.205
Since this is a probability question you need to have a sample space. In this situation we have already figured out the sample space in the previous question which is 20.5% or 0.205. We also need to have a number of favorable outcomes, In this case it is defective chips from factory A which is 0.175. To get the answer we divide the number of favourable outcomes by the sample space.
A box is randomly selected and a coin is chosen
a) The probability of a nickel.
P(N) = P(1N) + P(2N) + P(3N)
=1/9 + 1/9 + 1/4
=(4+4+9) / 36
In this question we just add up all the probabilities for getting nickels.
b) If a dime was chosen, what is the probability it came from box 3?
P(3(given that)D) = P(3D) / (P(1D)+P(2D)+P93D)
= (1/12) / (2/9 +2/9+ 1/12)
Since this is a probability question, we need to have a sample space. In this case the sample space is the probability of all dimes in all three boxes. The number of favorable outcome is the probability of chosing a dime from box three. So to get the answer you just divide the number of favourable outcomes by the sample space.
Well I guess that about wraps up my scribe. And oh yeah I almost forgot the next scribe..