Question 1010360
 
Question:
You have 18 cookies in a box. 5 are oatmeal, 4 are sugar, 3 are peanut butter, and 6 are butter. Show your calculations in answering the questions. 
A. Suppose you randomly choose a cookie, replace it, and randomly choose another. What is the probability that either the first cookie is butter or the second is oatmeal?
B. Suppose instead that you randomly choose a cookie, but you do not replace it. You then randomly choose another. What is the probability that either the first cookie is butter or the second cookie is oatmeal.
 
Solution:
 
Given: 18 cookies, 5 oatmeal (O), 4 sugar (S), 3 peanut butter (P) and 6 butter (B).
Let X be the event of drawing any other cookie.

A. P(BX or XO) with replacement
Use a contingency table (multiplied by 18^2=324).
   B  O  X
B 36 30 42
O 30 25 35
X 42 35 49

The entries are obtained by 
P(BB)=(6/18)(6/18)=36/324
P(BO)=(6/18)(5/18)=30/324
and so on.

P(BX or XO)
=(36+30+42)/324+(30+25+35)/324-30/324  (P(BO) was counted twice)
=14/27
alternatively,
P(BX or XO)
=P(B)+P(O)-P(BO)
=6/18+5/18-30/324
=14/27


B. P(BX or XO) Without replacement
Again, use a contingency table (multiplied by 18*17=306).
   B  O  X
B 30 30 42
O 30 20 35
X 42 35 42
The above table is obtained by
P(BB)=(6/18)(5/17)=30/306
P(BO)=(6/18)(5/17)=30/306
...
P(XX)=(7/18)(6/17)=42/306
So
P(BX or XO)
=(30+30+42)/306+(30+20+35)/306-30/306
=(102+85-30)/306
=157/306

Alternatively, 
P(BX or XO)
=P(B)+P(OO)+P(XO)
=6/18+(5/18)(4/17)+(7/18)(5/17)
=157/306