Unit 7: Probability

At least 1 problems

A manufacturer of processing chips knows that 2%, percent of its chips are defective in some way. Suppose an inspector randomly selects 4 chips for an inspection. Assuming the chips are independent, what is the probability that at least one of the selected chips is defective? Lets break this problem up into smaller pieces to understand the strategy behind solving it. Find the probability that a randomly selected chip is NOT defective: P (defective) = 0.02 P (not defective) = 0.98 Find the probability that all 4 chips are NOT defective: P(all 4 NOT defective) = (0.98)^4 P(all 4 NOT defective) = 0.922 Find the probability that 1 chip is defective: P(at least 1 defective) = 1 - P(all NOT defective) P(at least 1 defective) = 1 - (0.98)^4 P(at least 1 defective) = 1 - 0.922 P(at least 1 defective) = 0.078 NOTES: - (X)^the amt of sample. If we took 5 chips instead of 4, it would've been (0.98)^5 - Formula: P(at least one success) = 1 - P(all fail) P(at least one fail) = 1 - P(all success) - Big numbers are better (use 0.85 rather than 0.15)

General multiplication (independent) example

Ana is a teacher who plays a review game with her class. The game involves writing each student's name on an identical slip of paper and selecting students at random. her class is made up of 12 9th graders, 9 10th graders and 7 11th graders. Suppose that Ana picks a name, replaces it, and picks a name again. What is the probability that NEITHER of the students selected are 9th graders? There are 28 total students, 16 of which are not 9th graders. so, both picks are 16/28. P(not 9th grader) = (16/28) * (16/28) P(not 9th grader) = 256/784 P(not 9th grader) = 0.33

Probability w/ Venn Diagrams

Example: We have a shuffled deck of cards. (52, with 4 suits being the heart/club/clover/diamond and 13 in each suit) Well, what's the probability of drawing a jack? it'd be: 4 (jacks)/52 (cards) = 1/13. The probability of a heart is: 13(hearts)/52(cards) = 1/4 Probability of a jack of hearts is only 1/52, bc there is only one card like that & finally, the probability of EITHER a jack or a heart is 16/52 = 4/13 bc there are 3 jacks (thw one of hearts grouped with hearts) and 13 hearts, 3+13 is 16

Addition rule for probability

For any two events {a} and {b}, P(A or B) = P(A) + P(B) - P(A and B).

How to use the addition rule.

Let's say it's a problem with a venn diagram that has 20% no preference 10% only bought fries. 15% only bought burgers 55% bought both burgers and fries and we want to find P(burgers OR fries) 15% + 55% + 10% = 80%, or 0.8 P(burgers OR fries) = 0.8 ---> If it has no venn diagram Let's say a hospital tracks the day of birth that each baby was born, and if it was scheduled or unscheduled. We want to find P(Tuesday or Friday) 90 (total tuesday babies) + 70 (total friday babies) divided by 500 (total babies over week) Which comes out to 0.32, so P(Tuesday or Friday) = 0.32

General multiplication (dependent)

Let's say we have a crafting competition, with 2 finalists named Maya & Doug. For the final round, they need to pick a random card (w/out replacement) for their star material. The options are silk, fleece, plastic, wood, leather and felt. Both maya and doug want silk, and maya will draw first. What is the probability that neither one of them gets silk? Let's set up some P()s. MNS = Maya no silk DNS = Doug no silk P(MNS and DNS) = P(MNS) * P(MNS | DNS) There are 6 options for maya, 5 of which are not silk, so P(MNS) = 5/6 P(MNS and DNS) = (5/6) * P(MNS | DNS) There are only 5 options for Doug, 4 of which are not silk, so P(DNS) = 4/5. P(MNS and DNS) = (5/6) * (4/5) P(MNS and DNS) = 4/6 or 2/3

General multiplication (independent)

Let's say we have a crafting competition, with 2 finalists named Maya & Doug. For the final round, they need to spin a wheel that determines their star material. The wheel has silk, fleece, plastic, wood, leather and felt. Both maya and doug want silk, and maya will spin first. What is the probability that neither one of them gets silk? Let's set up some P()s. MNS = Maya no silk DNS = Doug no silk P(MNS and DNS) = P(MNS) * P(DNS | MNS) MNS and DNS are independent variables (they don't affect each other), so P(DNS | MNS) is the same as P(DNS) P(MNS and DNS) = P(MNS) * P(DNS) There are 6 total options on the wheel, 5 of which entail that Maya does not get silk. So, 5/6. P(MNS and DNS) = (5/6) * (5/6) P(MNS and DNS) = 25/36

General multiplication formula

P(A and B) = P(A) * P(B | A)

Conditional Probability w/ P()s

P(Bike) = 0.14 P(Crossing guard) = 0.48 P(Bike & Crossing Guard) = 0.12 P(bike | guard) = 0.25 [P(guard & bike)] / [P(guard)] = (0.12)/(0.48) = 1/4 P(guard | bike) = [P(guard & bike)] / [P(bike)] = (0.12)/(0.14) = 6/7

Reading Conditional Probability

P(male | fly) = X is the same as "X% of fliers are male" P(fly | male) = Y is the same as "Y% of males chose to fly"

Conditional Probability (2-way table)

The probability of a person choosing to fly is 0.38 38 fliers/100 total = 0.38 The probability someone was male is 0.48 48 boys/100 total = 0.48 The probability someone was male, given that they chose to fly as well. P(male | fly) = 0.68 P(male | fly) = 26 (ppl who both wanted to fly & were boys)/38 total fliers. P(male | fly) = 26/38 = 0.68 The probability someone was male, given that they chose to fly as well. P(fly | male) = 0.54 P(fly | male) = 26 (ppl who both wanted to fly & were boys)/48 total males. P(fly | male) = 26/48 = 0.54

Interpreting results of simulations

The probability that Maria succeeds at any given free-throw is 75%. She was curious how many free-throws she can expect to succeed in a sample of 10 free-throws. She simulated 25 samples of 10 free-throws where each free-throw had a 0.75 probability of being a success. Maria counted how many free-throws were successes in each simulated sample. Here are her results: 5: 0 successes 6: 5 successes 7: 8 successes 8: 6 successes 9: 6 successes 10: 1 success We need to use her results to estimate the probability that she succeeds at fewer than 7 free-throws in a sample of 10 free-throws. First, find out how many successes were less than 7 (out of 25), which is 5. now, we divide 5 by 25 (5/25, formula being number/total attempts) and simplify into a decimal, which comes out to 0.2

General multiplication (dependent) example

There are 50 workers onsite at a large construction project. Six of them are left-handed, and the rest are not. Suppose that the project manager randomly selects 2 workers (without replacement). What is the probability that NEITHER of the workers selected are left-handed? There are total 50, and 6 are left-handed, which means 44 are not left handed. The first pick is 44/50. The second pick is done without replacement, making it dependent and basically subtracting 1 from the top and bottom. the second pick is 43/49 P(neither left handed) = (44/50) * (43/49) P(neither left handed) = 946/1225 P(neither left handed) = 0.772

Mutually Exclusive events

events that cannot happen at the same time

Tip for general multiplication

try to logic it first.