For example: a class to learn a total of 100, of which 30% of the high number of hanging branches, wires hanging branches on behalf of 25%, then the high number of hanging branches of the students, families of algebraic probability is linked to how much? (Note: the high number and algebra while hanging branches, what is the probability there is a difference, the difference is that one is at the same time the entire class of hanging branches rate, which is the sample space is the entire class 100 students, while the sample space title is high number of hanging branches)
total sample space is: s = 100
probability a of the event: the high number of hanging branches probability = 30%
probability of B events: line represents hanging branches probability = 25%
in the case of hanging branches of the a, B event the probability is the conditional probability P (B | a) =
\ ({P \ left (B \ left |. A \ left) = \ frac {{P \ left (AB \ right)}} {{P \ left (A \ right)}} \ right \ right \ right. .} \)
i.e. the intersection of a and B divided by the space a sample points.
Examples: there are provided six ball box red, four white. Extraction without replacement, each either take a total extracted twice.
1) to take the first known white ball, requires computing a second probability to red ball.
2) taken to find the first white ball, the second probability to get red ball.
A hypothesis for the event to take the red ball, "take the red ball" for the first time as A1, the second pumping the ball to A2. Compared with the corresponding event A "to get white ball"
probability 1 in question to take the red ball first, then the remaining nine balls, taken in red ball 9, namely:
\ ([{P \ left (\ text {} A \ mathop {{ }} \ nolimits _ {{2}} \ left | \ overline {A} \ mathop {{}} \ nolimits _ {{1}} \ left) = \ frac {{6}} { {9}} = \ frac {
{2}} {{3}} \ right \ right \ right} \)... the difference between the two problems is a problem: a problem to a "known", is described conditions . question 2 is calculated Common step calculation.
\ ({\ the begin {Array} {* {20 is} {L}} {P \ left (\ text {} \ overline {a} \ mathop {{}} \ nolimits _ {{1}} A \ mathop {{}} \ nolimits _ {{2}} \ left) = P \ left (\ text {} \ overline {A} \ mathop {{}} \ nolimits _ {{1} } \ left) \ times P \ left (\ text {} A \ mathop {{}} \ nolimits _ {{2}} \ left | \ overline {A} \ mathop {{}} \ nolimits _ {{1}} \ left) = \ frac {{4 }} {{10}} \ times \ frac {{2}} {{3}} = \ frac {{4}} {{15}} \ right. \ right. \ right . \ right. \ right. \ right. \ right.} \\ {\ text {} \ text {} \ text {}} \ end {array}} \)
Example 2: Suppose there are three lottery tickets, of which only one is winning the lottery, the existing three students in turn drawn without replacement, ask a classmate finally winning probability is smaller than the other students.
Provided X = did not win, y = winning
sample space s = {yxx, xyx, xxy }, only three cases. Finally, a probability of winning is 1/3
but is known to the students did not win the first one, then the rest of the sample space is 2, the probability of winning this time last one is a 1/2
Nature of the conditional probability of
1) a non-negative: For each event B, there is P (B | A)> 0
2) specifications: for the inevitable event s, there is P (S | A) = 1
. 3) may be added to the column of: Let B1, B2 ... pairwise mutually exclusive events, there
\ ({\ begin {array} {* {20} {l}} {P \ left (\ mathop {{\ mathop {{\ cup}} \ limits ^ {{\ infty}}}} \ limits _ {{i = 1}} ^ {{} } B \ mathop {{}} \ nolimits _ {{i}} \ left | A \ left) = {\ mathop {\ sum} \ limits _ {{i = 1}} ^ {{ \ infty}} {P \ left (B \ mathop {{}...} \ nolimits _ {{i}} \ left | A \ right) \ right}} \ right \ right \ right} \\ {\ text. {} \ text {} \ text
{}} \ end {array}} \) and the total probability of all events and events in = the conditional probability of occurrence and.
Multiplication formula
\ ({\ begin {array} {* {20} {l}} {P \ left (AB \ left) = P \ left (B \ left | A \ left) \ times P \ left (A \ right ) \ right. \ right. \
right. \ right. \ right.} \\ {\ text {} \ text {} \ text {}} \ end {array}} \) be appreciated that as the probability of simultaneous AB = B probability of occurrence probability of the entire sample space a * occurs when the condition a occurs.
The type can be generalized to plot events more events. For example: Let A, B, C for the event, and P (AB)> 0, there is P (ABC) = P (C | AB) P (B | A) P (A)
\({\begin{array}{*{20}{l}} {P \left( AB \left) =P \left( B \left| A \left) \times P \left( A \right) \right. \right. \right. \right. \right. }\\ {P \left( B \left) =0.4,P \left( A+B \left) =0.5,\text{求}\text{ }P \left( A \left| \overline {B} \left) \text{ }\text{ }\right. \right. \right. \right. \right. \right. \right. }\\ {P \left( A \left) =1-0.5=0.5\right. \right. }\\ {P \left( \overline {B} \left) =1-0.4=0\text{ }.6\right. \right. }\\ {P \left( A \left| \overline {B} \left) =\frac{{P \left( A \overline {B} \right) }}{{P \left( \overline {B} \right) }}=\frac{{0.1}}{{0.6}}=\frac{{1}}{{6}}\right. \right. \right. }\\ {\text{ }}\\ {} \end{array}}\)
Figure understood
\ ({P \ left (A \ overline {B} \ right)} \) = A and B of intersection of the opposite event, i.e. 0.1
第二种办法：公式推导
\({P \left( A \left| \overline {B} \left) =1-P \left( \overline {A} \left| \overline {B} \left) =1-\frac{{1-P \left( A \cup B \right) }}{{1-P \left( B \right) }}=1-\frac{{1-0.5}}{{1-0.4}}=\frac{{1}}{{6}}\right. \right. \right. \right. \right. \right. }\)
Example 2: a savings card password of a total of six numbers, each digit can choose one from 0-9, withdraw money in the bank ATMs someone, forget the last digit password, seeking
1) according to any of the last digit, no more than 2 times to press on probability.
Press to set up events for the A, for the first time by A1, A2 of the secondary by
no more than 2 times the probability of the press, according to the first or second pair are regarded by no more than 2 times by right.
2) If he remembers the password last one is an even number, no more than 2 times to press probability of the
above it is the same, but the sample space becomes {0,2,4,6,8}, a total of five words
\ ({{P \ left ( \ mathop {{A}} \ nolimits _ {{1}} \ cup \ overline {\ mathop {{A}} \ nolimits _ {{1}}} A \ mathop {{}} \ nolimits _ {{2}} \ left) = \ frac {{1}} {{5}} + \ frac {{4}} {{5}} \ times \ frac {{1}} {{4}} = \ frac {{2}} { {5}} \ right. \ right.}} \)
Example 3: A Course class failure rate of 4%, while the students will pass 25% of students can wait A, find the probability of A to give the student.
Event A: have A, event B: pass
known:
\ ({\ Array the begin {} {} {L * 20 is {{}} A \ Subset B, AB = A, P \ left (AB \ left) = P \ left (A \ right) \ right. \ right.} \\ {P \ left (B \ left) = 1-4 \ text {%} = 96 \ text {%} \ right. \ right.} \\ {P \ left (A \ left | B \ left) = 25 \ text {%} \ right \ right \ right...} \\ {P \ left (AB \ left) = P \ left (B \ left) \ times P \ left (A \ left | B \ left) = 96 \ text {%} \ times 25 \ text {%} = 0.24 \ right \ right \ right \ right \ right \ right \...... right.} \ end {array} } \)