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Conditional probability exercise

  • #1
I have some problems getting conditional probability right... Does this look like it should?

Homework Statement


Assume that there are bags of tulip bulbs in the basement, ant that they contain 25 bulbs each. yellow bags contain 20 yellow tulips and 5 red tuplips, and red bags contain 15 red and 10 yellow tulips. 60% of the bags in the basement are yellow, the others are red. One bulb is chosen at random from a random bag in the basement, and then planted

a) what is the probabilit that the tulip turns out yellow?
b) given that the tulip turns out yellow, what is the probability that it came from a yellow bag?

Homework Equations


Let RB be redbag, YB yellowbag, RT red tulip and YT yellow tulip. Then as far as I can read from this exercise I have the following information:

P(YB)=0.6
P(RB)=0.4
P(YT|YB)=20/25=4/5
P(RT|YB)=1/5
P(YT|RB)=10/25=2/5
P(RT|RB)=3/5


The Attempt at a Solution


a) [itex] P(YT)=P(YB\cap YT)+P(RB\cap YT)=(0.6*4/5)+(0.4*2/5)=16/25 [/itex]
b) [itex] P(YB|YT)=P(YB \cap YT)/P(YT) = (0,6*4/5)/(16/25)=3/4 [/itex]

It is particularly this last one I am unsure about.
 

Answers and Replies

  • #2
LCKurtz
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I get the same answers.
 
  • #3
vela
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I have some problems getting conditional probability right... Does this look like it should?

Homework Statement


Assume that there are bags of tulip bulbs in the basement, ant that they contain 25 bulbs each. yellow bags contain 20 yellow tulips and 5 red tuplips, and red bags contain 15 red and 10 yellow tulips. 60% of the bags in the basement are yellow, the others are red. One bulb is chosen at random from a random bag in the basement, and then planted

a) what is the probabilit that the tulip turns out yellow?
b) given that the tulip turns out yellow, what is the probability that it came from a yellow bag?

Homework Equations


Let RB be redbag, YB yellowbag, RT red tulip and YT yellow tulip. Then as far as I can read from this exercise I have the following information:

P(YB)=0.6
P(RB)=0.4
P(YT|YB)=20/25=4/5
P(RT|YB)=1/5
P(YT|RB)=10/25=2/5
P(RT|RB)=3/5


The Attempt at a Solution


a) [itex] P(YT)=P(YB\cap YT)+P(RB\cap YT)=(0.6*4/5)+(0.4*2/5)=16/25 [/itex]
b) [itex] P(YB|YT)=P(YB \cap YT)/P(YT) = (0,6*4/5)/(16/25)=3/4 [/itex]

It is particularly this last one I am unsure about.
If you calculate the probabilities of the various outcomes, you find
\begin{align*}
P(YB \cap YT) &= \frac{12}{25} \\
P(RB \cap YT) &= \frac{4}{25} \\
P(YB \cap RT) &= \frac{3}{25} \\
P(RB \cap RT) &= \frac{6}{25}.
\end{align*} This grid represents the sample space
Code:
Yy Yy Yy Yy Yr 
Yy Yy Yy Yy Yr 
Yy Yy Yy Yy Yr 
Ry Ry Rr Rr Rr 
Ry Ry Rr Rr Rr
where the capital letter represents the color of the bag, and the lowercase letter represents the color of the tulip.

On average, if you plant 25 tulips, you'll get 16 yellow tulips where 12 times you'll have drawn a yellow tulip from a yellow bag and 4 times you'll have drawn a yellow tulip from a red bag. If you know you have a yellow tulip — that is, the probability of a yellow tulip is 1 — you are restricting your sample space to those 16 outcomes which result in a yellow tulip, that is, the sixteen outcomes bolded in the grid below:
Code:
[b]Yy Yy Yy Yy[/b] Yr 
[B]Yy Yy Yy Yy[/B] Yr 
[B]Yy Yy Yy Yy[/B] Yr 
[B]Ry Ry[/B] Rr Rr Rr 
[B]Ry Ry[/B] Rr Rr Rr
Of those 16, a yellow tulip will have come from a yellow bag on average 12 times, so the probability of drawing from a yellow bag is 12/16.

With a conditional probability, you're simply renormalizing your probabilities taking into account the given information.
 

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