Question

An experiment consists of reaching into a hat with six chips of identical size and shape, except two each are marked with 2, 3, or 4. Two chips are grabbed and the total of the chips is determined. (Hint: there are 6 choose 2ways to grab two in the Space, S) Some possible Events are E =( the total is odd )and F {6 8} a) Build a PMF (an assignment, because an actual function is not possible). b) Find P(E)  . c) Find P (E if F) . 4pts

Answer 1

Answer:

Given Data

Q1) We are given the events here as:
E = { Total is odd }

F = { 6 <= Total <= 8 }

a) There are a total of 6c2 = 15 ways to select 2 tiles here. Therefore the PMF for the sum of two tiles is computed here as:
P(X = 2 + 2 = 4) = 1/15
P(X = 2 + 3 = 5) = 2*2 / 15 = 4/15
P(X = 2 + 4 = 6 or 3 + 3 = 6) = (2*2 + 1)/15 = 5/15
P(X = 4 + 4 = 8) = 1/15
P(X = 3 + 4 = 7) = 2*2 / 15 = 4/15

Therefore the PMF here is given as:
P(X = 4) = 1/15
P(X = 5) = 4/15
P(X = 6) = 5/15
P(X = 7) = 4/15
P(X = 8) = 1/15

b) The probability here is computed as:
P(E) = P(Total is odd) = P(X = 5 or 7) = (4/15) + (4/15) = 8/15
Therefore 8/15 is the required probability here.

c) P(E | F) is computed using Bayes theorem as:
= P(E and F) / P(F)

= P(X = 7) / P(X = 6, 7 or 8)

= 4 / (5 +4 + 1)

= 0.4

Therefore 0.4 is the required probability here.

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