Thread replies: 2
Thread images: 1
Thread images: 1
Anonymous
College Statistics Help 2017-06-12 12:14:10 Post No. 330310
[Report] Image search: [Google]
College Statistics Help 2017-06-12 12:14:10 Post No. 330310
[Report] Image search: [Google]
File: sensible chuuckle.png (393KB, 611x607px) Image search:
[Google]
393KB, 611x607px
The Question is
Let X, Y , Z be coins in a box. Suppose X is a fair coin, Y is a two-headed coin, and Z is weighted so that the probability of heads is 1/3. A coin is selected at random and is tossed.
A)If heads appears, find the probability that it came from the fair coin, X.
B)If tails appears, find the probability that it came from coin Z.
The answer is supposed to be a) "P(X|H) = 3/11" and b) "P(Z|T) = 4/7" But I don't know how they get there.
The formula for this sort of thing usually is: P(A|B)=P(A and B)/P(A) but that isn't working at all for this, at least not the way I'm doing it. I have H=11/18 which the answer sheet also says is right but I still don't know how to get to the answer.
>>
>>330310
There are only six possible scenarios, so just enumerate the lot of them
X1=>H(1/6)
X2=>T(1/6)
Y=>H(1/3)
Z1=>H(1/9)
Z2=>T(2/9)
Total P of getting a H: 1/9+1/3+1/6 = 6/54+18/54+9/54=33/54=11/18
P of getting a H from X = 1/6 = 3/18
(3/18)/(11/18) = 3/11
Thread posts: 2
Thread images: 1
Thread images: 1