More Poisson & Hypergeometric Probability Distribution

1. Suppose that a randomly selected person contracts no more than two colds in a given year. What is the probability that person takes/took Vitamin C supplements?

   P(C|X<2)= P(C ∩ X ≤ 2)/P(X ≤ 2)= (0.7*0.423)/0.3336=0.8876

Continuous inspection of electrolytic tin plates yields on average 0.2 imperfections per minute.

In light of the fact that we’re making the following assumptions:

• Assume that this rate remains constant in time.
• Assume we don’t expect to find 2 imperfections simultaneously (i.e. events are rare).
• Assume that the arrival of imperfections occurs independently.

given lambda = 2 imperfections a minute

assuming poisson proccess

1. the probability of one imperfection in three minutes

   X*~poisson(.2*3) -> because .2 per minute so multiply this by 3 for 3 mintues

   X*~poisson(.6)

   P(X*=1) = (.6¹*e^-.6)/!1 = .329

2. the probability of at least two imperfections in five minutes

   X*~poisson(.2*5)

   X*~poisson(1)

   P(X*=2) = 1 - poisson(x*<2) = .264

3. the probability of at most one imperfection in .25 hours is…

   P(X* <= 1) = P(X*=0) + P(X*=1)

Recall the example where we had 1,000,000 people and 90% played soccer and we selected 2 people at random (Messi and Ronaldo).

P(Messi and Ronaldo play soccer)

=P(Messi Plays) * P(Ronaldo Plays | Messi Plays)

=900,000/1,000,000*899,999/999,999

=*0.80999991*

But since, we only selected 2 people (<10% of 1 million), we can just say:

P(Messi plays)*P(Ronaldo plays|Messi plays)

= P(Messi plays)*P(Ronaldo plays) -> Treat them as independent trials…

=900,000/1,000,000 * 900,000/1,000,000

=0.9*0.9

=*0.81*

We could do this because the number of trials was < 10% of the total size of the population.

In today’s examples, we will have the opposite. The number of trials will be > 10% of the total size of the population.

Instead of the Binomial Distribution, we’ll use the Hypergeometric Distribution.

Twelve refrigerators have been returned to the distributor because of a high-pitched oscillating noise. Suppose you know that 4 of the 12 have a defective compressor and the rest have less-serious problems. Someone else (that doesn’t know this information) selects 6 refrigerators at random for problem identification. Let X be the number of those found with a defective compressor.

x = # of deffective compressors (in my selection of 6 fridges)

x is hypergeometric b/c

• success = defective & fail = non-defective
• 6 trials
• selected 6/12 = 50% > 10% of the population (no independance)

Sample space of X 4 D’s, 8 N’s, selecting 6

X = {0, 1, 2, 3, 4, 5, 6}

X = # Defective in sample of 6

we could do multiplication rule but without independance that is too much work

P(X = 3)

D1 D2 D3 d4 N1 N2 N3 N4 N5 N6 N7 N8

theres 12 and we need 6

order does not matter so…

(12 choose 6) = 924

Template

D D D N N N

Sample D1 D2 D3 N1 N2 N3

4 3 2/3! 8 7 6/3!

= (4 choose 3) * (8 choose 3) = 234

P(X=3) =((4 choose 3) * (8 choose 3))/(12 choose 6) = .2424