Question

Out of 52 playing cards one is lost from the remaining cards . 2cards where chosed and were found spaid find the probablity that the missing card is a spaid?

Answer 1

yes the same.

Answer 2

i think this is going to take a bit of work, and also need baye's formula.
if i understand the question, some card is missing, you don't know what
you pick two cards
both are spades
question is "what is the probability that the missing card is a spade?"
is that right?

Answer 3

exactly

Answer 4

then unless i miss my guess, you are going to use baye's formula for this. does that sound right?

Answer 5

yes conditional probability!! = baye's formula

Answer 6

so if we put A= both cards chosen are spades
B= the missing card is a spade
then we want
\[P(B|A)\] and by baye's we get that
\[P(B|A)=\frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|B^c)P(B^c)}\] so we have a bit of computing to do

Answer 7

\[P(B)=\frac{1}{4}\] is easy, as is
\[P(B^c)=\frac{3}{4}\]

Answer 8

now for
\[P(A|B)\] we assume that the missing card is a spade, and compute the probability that we draw two spades in a row. since the missing card is a spade there are 12 spades left in the deck and 51 cards in the deck so the probability of drawing two spades under these circumstances is
\[\frac{12}{51}\times\frac{11}{50}\]

Answer 9

as for
\[P(A|B^c)\] you know that the missing card was NOT as spade, so there are 13 spades in the deck of 51 cards and the probability you draw two spades in a row is now
\[\frac{13}{51}\times \frac{12}{50}\]

Answer 10

now we have all the numbers we need to compute the left hand side of
\[P(B|A)=\frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|B^c)P(B^c)}\] which i will let you do