SOLUTION: 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

Question 1010360: 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.
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

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
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