• April 29th 2008, 08:00 AM
matty888
Suppose there are two bowls full of cookies.Bowl 1 has 20 chocolate chip and 30 plain cookies,while bowl 2 has 25 of each. Fred picks a bowl at random and then picks a cookie at random
i) What is the probability that Fred picks a plain cookie?
ii) If the cookie Fred picked is a plain one,what is the probability that it was from bowl 1?
• April 29th 2008, 08:43 AM
janvdl
Quote:

Originally Posted by matty888
Suppose there are two bowls full of cookies.Bowl 1 has 20 chocolate chip and 30 plain cookies,while bowl 2 has 25 of each. Fred picks a bowl at random and then picks a cookie at random
i) What is the probability that Fred picks a plain cookie?
ii) If the cookie Fred picked is a plain one,what is the probability that it was from bowl 1?

Bowl 1 has 50% of the cookies and so too does Bowl 2.

Bowl 1 contains 40% choc chip and 60% plain.

Bowl 2 contains 50% choc chip and 50% plain.

Use Bayes' Rule. (Dont you just love this rule? :D )

$
P(B_{1} | P) = \frac{ P(P | B_{1}) P(B_{1})} {P(P | B_{1}) P(B_{1}) + P(P | B_{2}) P(B_{2})}$

$
P(B_{1} | P) = \frac{ (0,6)(0,5) }{ (0,6)(0,5) + (0,5)(0,5) }$

$P(B_{1} | P) = \frac{6}{11}$
• April 29th 2008, 08:54 AM
matty888
Thanks
Thanks for the answer,would I be right by saying that the answer to part i) is
11/20
• April 29th 2008, 08:59 AM
yup thats correct
• April 29th 2008, 09:03 AM
janvdl
Quote:

Originally Posted by matty888
Suppose there are two bowls full of cookies.Bowl 1 has 20 chocolate chip and 30 plain cookies,while bowl 2 has 25 of each. Fred picks a bowl at random and then picks a cookie at random
i) What is the probability that Fred picks a plain cookie?
ii) If the cookie Fred picked is a plain one,what is the probability that it was from bowl 1?

Quote:

Originally Posted by matty888
Thanks for the answer,would I be right by saying that the answer to part i) is
11/20

Sorry, I only did part 2... My apologies.

For part 1:

$P(Plain) = P(P | B_{1})P(B_{1}) + P(P | B_{2})P(B_{2})$

$P(Plain) = 0,6(0,5) + 0,5(0,5)$

$P(Plain) = \frac{11}{20}$
• April 29th 2008, 10:45 AM
Soroban
Hello, Matty!

A slightly different approach . . .

Quote:

Suppose there are two bowls full of cookies.
Bowl #1 has 20 chocolate chip and 30 plain cookies; Bowl #2 has 25 of each.

Fred picks a bowl at random and then picks a cookie at random.

a) What is the probability that Fred picks a plain cookie?

$P(\text{Bowl \#1} \:\wedge\text{ Plain}) \:=\:\left(\frac{1}{2} \right)\left(\frac{30}{50}\right) \:=\:\frac{3}{10}$

$P(\text{Bowl \#2} \:\wedge\text{ Plain}) \:=\:\left(\frac{1}{2}\right)\left(\frac{25}{50}\r ight) \:=\:\frac{1}{4}$

. . Therefore: . $P(\text{Plain}) \;=\;\frac{3}{10} + \frac{1}{4} \;=\;\boxed{\frac{11}{20}}$

Quote:

bi) If the cookie Fred picked is a plain one,
what is the probability that it was from Bowl #1?

Bayes' Theorem: . $P(A\,|\,B) \;=\;\frac{P(A\,\wedge B)}{P(B)}$

We have: . $P(\text{Bowl \#1 }|\text{ Plain}) \;=\;\frac{P(\text{Bowl \#1 }\wedge\text{ Plain})}{P(\text{Plain})} \;=\;\frac{\dfrac{3}{10}}{\dfrac{11}{20}} \;=\;\boxed{\frac{6}{11}}$

• April 29th 2008, 10:51 AM
matty888
Hello and thanks vey much Soroban.Thanks janvdl and Danshader too!!!(Wink)(Rofl)