conditional probability question involving cards.

Question

given that 13 cards are dealt from a deck of 52 cards. ​(a) What is the probability that the ace of spades is one of the 13 ​cards?

$\frac{1}{4}$

​(b) Suppose one of the 13 cards is chosen at random and found not to be the ace of spades. What is the probability that none of the 13 cards is the ace of​ spades?

i get .760 the correct answer is .765 it must be using conditional probabilty can someone explain how?

​(c) Suppose the experiment in part​ (b) is repeated a total of 10 times​ (replacing the card looked at each​ time), and the ace of spades is not seen. What is the probability that the ace of spades actually is one of the 13 ​cards?

i dont see why this is .130

Answer 1

For part $b$:

There are $12$ unexposed cards in your hand, and $39$ unexposed cards remaining in the deck. The probability that the designated card is one of the $39$ is then $$\frac {39}{51}\approx .7647$$

For $c$:

This calls for Bayes Theorem. As a prior, we think the designated card has a $\frac 14$ chance of being in your hand, and a $\frac 34$ chance of being in the rest of the deck. If it is in your hand, these ten tests have a $\left( \frac {12}{13}\right)^{10}$ chance of failing to turn it up. Of course if it is not in your hand then the tests are guaranteed to fail. Bayes' then tells us that the revised estimate of the probability is $$\frac {.25\times \left( \frac {12}{13}\right)^{10}}{.25\times \left( \frac {12}{13}\right)^{10}+.75 \times 1}\approx .1302$$

Answer 2

Here is how Bayes may join the lulu solution

a)

$ \frac{1}{ 1\over 13} = \frac {P(AS\ in\ hand\ |\ success) } {P(success\ |\ AS\ in\ hand)} = \frac {P(AS\ in\ hand)}{P(success)} = \frac {x} {1\over 52}$

and we get the first 1/4

b) reusing 1/4,

$ \frac{x}{ 12\over 13} = \frac {P(AS\ in\ hand\ |\ failure) } {P(failure\ |\ AS\ in\ hand)} = \frac {P(AS\ in\ hand)}{P(failure)} = \frac {1\over 4} {51\over 52}$

we obtain the 1 - 12/51

c) in fact we are interested in

$ \frac{x} { ( {12 \over 13} ) ^{10} } = \frac {P(AS\ in\ hand\ |\ 10\ failures) } {P(10\ failures\ |\ AS\ in\ hand)} = \frac {P(AS\ in\ hand)}{P(10\ failures)} = \frac {1\over 4} {y}$

where $ y = P(10\ failures) = P(10\ failures \cap AS\ in\ hand) + P(10\ failures \cap AS\ not\ in\ hand) $

hence the result

$ x = \frac{P(10\ failures \cap AS\ in\ hand)} {P(10\ failures)} $

In fact

a) Bayes is not needed

b) Bayes is not needed

c) Bayes in not needed.