Asked by Anonymous
Six cards-ace, king, queen of hearts; ace, king of diamonds; and king of clubs-are all placed face down a table. One card is to be selected at random. Find the probability:
a) of an ace, given that a diamond was turned
b) of an ace, given that a club was turned
c) of a heart, given that an ace was turned
d) of a club, given that a king was turned
e) Explain why it is true that P(A/C)=P(C/A)
After completing a through d, I thought I did them correctly unil I reached to e. Here are the answers I got: a) 2/5 b)2/5 c)1/5 d)1/5
a) of an ace, given that a diamond was turned
b) of an ace, given that a club was turned
c) of a heart, given that an ace was turned
d) of a club, given that a king was turned
e) Explain why it is true that P(A/C)=P(C/A)
After completing a through d, I thought I did them correctly unil I reached to e. Here are the answers I got: a) 2/5 b)2/5 c)1/5 d)1/5
Answers
Answered by
economyst
Hummm, I do not agree with your answers a-d.
a) given that a diamond was turned, it must have been an ace or king. So for a) I get 1/2.
b) given that a club was turned, it must have been a king. So for b) I get a zero probability.
Now then, notationally, does P(A/C) mean probability of an Ace given a club was turned. If so, P(A/C)=0, and P(C/A)=0
a) given that a diamond was turned, it must have been an ace or king. So for a) I get 1/2.
b) given that a club was turned, it must have been a king. So for b) I get a zero probability.
Now then, notationally, does P(A/C) mean probability of an Ace given a club was turned. If so, P(A/C)=0, and P(C/A)=0
Answered by
Anonymous
how is that so when there are five cards faced down?
Answered by
Anonymous
I get it. nevermind.
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