Methods to find probability II

Methods to find probability I

Tree diagram, tables

Conditional probability
$P(A|B)=P(AB)/P(B)$

Rule of total probability
$P(A)=P(A|B)+P(A|B^c)$
Sometimes P(A) itself is hard to find, we can use the rule of total probability to find P(A) indirectly.

Reduced sample space
"In the card game bridge, the 52 cards are dealt out equally to 4 players—called East, West, North, and South. If North and South have a total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades?"

Since we know that North and South have 8 spades among their 26 cards, East and West together have 26 cards, 5 of them being spades. Therefore, the probability that East has 3 of the 5 spades is given by
$\large \binom{5}{3} \binom{21}{10} / \binom{26}{13}$.

Note that this is actually a hypergeometric distribution, which we will introduce in our next post.
Further note that this problem looks very much like a conditional probability problem in the form "Given that ..., find the probability that ..." We will show you the calculations using conditional probability, but the method of reduced sample space would be far more straightforward.

Let B denote the event North and South have a total of 8 spades and 18 non-spades.
$\large P(B)=\binom{13}{8} \binom{39}{18}/ \binom{52}{26}\\$
Let A denote the event East has a total of 3 spades and 10 non-spades.
$\large P(AB)=\binom{13}{8} \binom{39}{18} \binom{5}{3} \binom{21}{10} / [\binom{52}{26}\binom{26}{13}]\\
\large P(A|B)=P(AB)/P(B)=\binom{5}{3} \binom{21}{10}/ \binom{26}{13}$

Bayes' formula
$P(A|B)=P(B|A)P(A)/P(B)$

Inclusion-exclusion principle
Let's say you toss a dice for four times. What is the probability that there is at least one 3 in the four tosses? We can make use of complement to simplify our calculations. The required probability is just $1-\text{P(no 3s)}=1-(5/6)^4=671/1296$. What about calculating the probability directly? We use the inclusion-exclusion principle. $P(\bigcup\limits_{i=1}^n A_i)=\sum\limits_{i=1}^nP(A_i)-\sum\limits_{i<j}P(A_iA_j)+\sum\limits_{i<j<k}P(A_iA_jA_k)-\cdots+(-1)^{n-1}P(\bigcap\limits_{i=1}^n A_i)$

Let $A_i$ denote the event ith dice is a 3.
Thus we need to find $P(\bigcup\limits_{i=1}^4 A_i)$. By the inclusion-exclusion principle, it is $\sum\limits_{i=1}^4 P(A_i)-\sum\limits_{1<i<j<4}P(A_iA_j)+\sum\limits_{i<j<k}P(A_iA_jA_k)-\cdots-P(A_1A_2A_3A_4)$
$=4(1/6)-6(1/6)^2+4(1/6)^3-(1/6)^4=671/1296$.

Permutations and combinations