Statistics and Probability - Statistical Methods - Notes | STT 200, Study notes of Data Analysis & Statistical Methods

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1 Week 3 -16-

In the classical model all outcomes are regarded as being equally likely.

P(event) = n(event) / n(S) an event is just a subset of the sample space S

P(event) = n(event) / n(S) an event is just a subset of the sample space take the ratio of favorable outcomes to total outcomes

total favorable cases 2 1 3 2 4 3

... 7 6 8 5 ... 12 1 e.g. P( 1 0 ) = 3 / 36 = 1 / 12 from 6 , 4 or 5 , 5 or 4 , 6.

type favorable aa 1 aA 2 AA 1 P( aa ] = 1 / 4 P( aA ) = 2 / 4 = 1 / 2 P( AA ) = 1 / 4 Dad Mom

Jack draws a bill first Jill draws second from the two bills then remaining

{ a, b, c } Jack first, Jill second draw BILLS not dollar amounts

{$ 1 , $ 1 , $ 5 }Jack first, Jill second from the two bills then remaining

{$ 1 , $ 1 , $ 5 } Jack first, Jill second from the two bills then remaining

{$ 1 , $ 1 , $ 5 }Jack first, Jill second from the two bills then remaining

draws without replacement from {R, R, R, B, B, G}

draws without replacement from {R, R, R, B, B, G} same

Suppose that each birth is independently placed into one of 365 days. The chance that all of a given number n of birthdays will differ is 364/365 3 63/365 … (366-n)/ 2nd misses first 3rd misses 1st and 2nd etc. This is around e^(- n (n-1)/730) ~ 1/2 for n = 23. That is, around 50% of the time there would be no shared birthdays among 23 persons. By complements, there is around 1/2 probability of at least one instance of same birthdays among 23 persons, and even greater in the real world where some days have more births.