- WELCOME TO A LESSON ON CONDITIONAL PROBABILITY.
AGAIN BEFORE WE START REMEMBER THE PROBABILITY OF AN EVENT
IS EQUAL TO THE FAVORABLE NUMBER OF OUTCOMES
COMPARED TO THE TOTAL NUMBER OF OUTCOMES
WHICH CAN BE EXPRESSED AS A FRACTION, DECIMAL OR PERCENTAGE.
CONDITIONAL PROBABILITY IS THE PROBABILITY OF AN EVENT E2,
OCCURRING GIVEN THAT EVENT E1 HAS ALREADY OCCURRED.
THIS IS HOW WE DENOTE THE SITUATION.
THE PROBABILITY OF E SUB 2 GIVEN E SUB 1
IS EQUAL TO THE PROBABILITY OF E SUB 2
AND E SUB 1 DIVIDED BY THE PROBABILITY OF E SUB 1.
NOW THERE'S ONE THING TO KEEP IN MIND HERE.
IT'S GREAT TO HAVE ALL THESE FORMULAS TO HELP US OUT
BUT IF WE CAN MAKE SENSE OF THE SITUATION,
THEN WE DON'T HAVE TO RELY ON THESE FORMULAS
AND WE CAN ACTUALLY UNDERSTAND EXACTLY WHAT'S HAPPENING.
SO FOR THIS FIRST PROBLEM
WE'LL TAKE A LOOK AT IT USING 2 DIFFERENT METHODS.
ONE, JUST THE BASIC PROBABILITY FORMULA
AND TWO, USING THE CONDITIONAL PROBABILITY FORMULA.
AND HERE'S THE PROBLEM.
SO IF YOU ROLL ONE SIX SIDED DIE,
WHAT IS THE PROBABILITY OF A 3
GIVEN YOU KNOW THE NUMBER IS ODD?
SO WE HAVE THE PROBABILITY OF A 3 GIVEN WE KNOW THAT IT'S ODD.
WELL IF WE THINK ABOUT IT LOGICALLY,
IF WE KNOW THAT THE NUMBER IS ALREADY ODD
THAT REDUCES THE TOTAL NUMBER OF OUTCOMES.
THERE ARE 3 NUMBERS FROM 1 TO 6 THAT ARE ODD, 1, 3 AND 5.
SO THERE'S 3 POSSIBLE OUTCOMES GIVEN THAT WE KNOW THAT IT'S ODD
AND THERE'S ONLY ONE ODD NUMBER THAT IS 3,
SO THE PROBABILITY WOULD BE 1/3.
NOW IF WE TAKE A LOOK AT THE CONDITIONAL PROBABILITY FORMULA,
OF COURSE WE SHOULD GET THE SAME ANSWER
BUT IT'S GOING TO LOOK A LITTLE BIT DIFFERENT.
LET'S SEE IF WE CAN APPLY THIS AND GET THE SAME ANSWER.
SO THIS WOULD BE EQUAL TO THE PROBABILITY OF ROLLING A 3
AND AN ODD DIVIDED BY THE PROBABILITY OF ROLLING AN ODD.
WELL THE PROBABILITY OF ROLLING AN 3 AND AN ODD,
THERE'S ONLY 1 NUMBER THAT'S A 3 AND ODD.
SO THAT WOULD BE 1/6 DIVIDED BY THE PROBABILITY
OF ROLLING AN ODD WOULD BE 3/6.
WELL, 1/6 DIVIDED BY 3/6 IS THE SAME AS 1/6 x 6/3
AND AS YOU CAN SEE THE RESULT IS THE SAME. WE HAVE 1/3.
NOW YOU MAY BE ASKING
WHY WOULD YOU WANT TO USE THIS FORMULA ANYWAY,
'CAUSE IT SEEMS A LOT MORE COMPLICATED.
BUT IN SOME CASES IT WILL BE APPROPRIATE
AS WE SEE IN THE NEXT PROBLEM.
AT P TOWN HIGH SCHOOL,
THE PROBABILITY THAT A STUDENT TAKES COMPUTER PROGRAMMING
AND SPANISH IS 0.15.
THE PROBABILITY THAT A STUDENT TAKES COMPUTER PROGRAMMING
IS 0.4. AND HERE'S THE QUESTION.
WHAT IS THE PROBABILITY THAT A STUDENT TAKES SPANISH
GIVEN THAT THE STUDENT IS TAKING COMPUTER PROGRAMMING.
SO WE WANT TO KNOW WHAT THE PROBABILITY IS
THAT A STUDENT WILL TAKE SPANISH GIVEN THAT WE ALREADY KNOW
THEY'RE TAKING COMPUTER PROGRAMMING.
THIS WILL EQUAL THE PROBABILITY OF A STUDENT TAKING SPANISH
AND COMPUTER PROGRAMMING
DIVIDED BY THE PROBABILITY
OF A STUDENT TAKING COMPUTER PROGRAMMING.
AGAIN REMEMBER THAT E SUB 1 IS THE EVENT THAT ALREADY OCCURRED.
BASED UPON THIS QUESTION
WE ALREADY KNOW THEY'RE ARE TAKING COMPUTER PROGRAMMING.
SO WE SEE COMPUTER PROGRAMMING HERE
AND ALSO IN THE DENOMINATOR.
SO THE PROBABILITY OF A STUDENT TAKING SPANISH
AND A COMPUTER PROGRAMMING IS 0.15
AND THE PROBABILITY OF A STUDENT
TAKING COMPUTER PROGRAMMING IS 0.4.
WE CAN GO AHEAD AND DIVIDE THIS, BUT IT'S EQUAL TO 0.375
WHICH COULD BE EXPRESSED AS 37.5%.
AND AGAIN WHAT THIS TELLS US IS IF WE KNOW A STUDENT
IS ALREADY TAKING COMPUTER PROGRAMMING,
THEY HAVE A 37.5% CHANCE THEY WOULD BE IN SPANISH AS WELL.
LET'S GO AHEAD AND TAKE A LOOK AT ANOTHER ONE.
HERE'S A SURVEY COMPLETED BY ADULTS WITH CHILDREN.
AND THEY EITHER HAD CHILDREN IN COLLEGE OR NOT IN COLLEGE
AND THEY WERE ASKED WHETHER THEY THOUGHT COLLEGE
WAS EITHER TOO EXPENSIVE, AFFORDABLE OR TOO CHEAP.
AND HERE'S THE QUESTION.
WHAT IS THE PROBABILITY
THAT A PERSON THINKS COLLEGE IS TOO EXPENSIVE
GIVEN THAT THEY HAVE A CHILD IN COLLEGE.
SO WE'RE ONLY CONCERNED ABOUT IF THE PERSON WAS IN THIS ROW,
WHAT'S THE PROBABILITY
THAT THEY WOULD THINK THAT COLLEGE WAS TOO EXPENSIVE?
SO LOOKING AT OUR FORMULA HERE,
IT'S GOING TO EQUAL THE PROBABILITY
THAT ADULT PARENTS THINK THAT COLLEGE IS TOO EXPENSIVE
AND THEY HAVE A CHILD IN COLLEGE WHICH IS RIGHT HERE AS 55%.
NOW I'LL WRITE THAT IN DECIMAL FORM AS 0.55
DIVIDED BY THE PROBABILITY OF EVENT 1,
WHICH IS THEY HAVE A CHILD IN COLLEGE.
SO THE PROBABILITY THEY HAVE A CHILD IN COLLEGE
WOULD BE THIS ENTIRE ROW,
WHICH SHOULD BE 55% + 4% OR 59% OR 0.59.
SO WE TAKE A LOOK AT THIS LOGICALLY.
IF WE'RE ONLY CONSIDERING THE PARENTS WITH A CHILD IN COLLEGE,
THAT WOULD BE THIS 59% AS WE SEE HERE.
AND THEN OF THE 59%, THIS IS A PORTION,
THIS IS THE PORTION THAT THOUGHT COLLEGE WAS TOO EXPENSIVE.
SO LET'S GO AHEAD AND DIVIDE THIS OUT, 0.55 DIVIDED BY 0.59.
SO IT COMES OUT TO ABOUT 0.932, OR IF WE WANT 93.2%.
HOPEFULLY THAT MAKES SENSE
BECAUSE OF THE PEOPLE THAT HAVE CHILDREN IN COLLEGE,
WE CAN SEE THAT MOST OF THEM THOUGHT IT WAS TOO EXPENSIVE
COMPARED TO A VERY SMALL PERCENTAGE
THAT THOUGHT IT WAS AFFORDABLE.
LET'S TAKE A LOOK AT A COUPLE OTHERS.
THESE TWO, I THINK IT'S GOING TO MAKE IT VERY CLEAR
THAT IF WE CAN MAKE SENSE OF THE PROBLEMS
WE DO NOT WANT TO JUST MEMORIZE
OR RELY ON SOME SPECIAL FORMULA FOR A CONDITIONAL PROBABILITY.
HERE WE HAVE A STANDARD DECK OF CARDS.
TWO CARDS ARE DRAWN WITHOUT REPLACEMENT IN SUCCESSION.
WHAT IS THE PROBABILITY THAT THE SECOND CARD DRAWN IS AN ACE
GIVEN THAT THE FIRST CARD WAS AN ACE.
AGAIN JUST THINKING OF PROBABILITY
AS THE COMPARISON OF FAVORABLE OUTCOMES
TO TOTAL OUTCOMES,
IF WE KNOW THE FIRST CARD IS AN ACE,
THAT LEAVES A TOTAL OF 51 CARDS.
NOW WE ALSO KNOW THAT THE FIRST CARD IS AN ACE
AND THERE ARE A TOTAL OF 4 ACES IN A DECK.
SO IF THE FIRST CARD IS AN ACE,
THAT ONLY LEAVES 3 ACES LEFT IN THE DECK
AND THEREFORE THE PROBABILITY OF DRAWING AN ACE
GIVEN THE FIRST CARD IS AN ACE WOULD BE 3/51.
AND THIS IS THE PROBABILITY THAT WE'RE LOOKING FOR
AND THAT'S ALL THERE IS TO THIS QUESTION.
SO EVEN THOUGH IT'S CONDITIONAL PROBABILITY,
IF WE COULD MAKE SENSE OF IT, IT CAN BE VERY STRAIGHTFORWARD.
FOR NUMBER 2, 2 CARDS ARE DRAWN WITHOUT REPLACEMENT,
WHAT IS THE PROBABILITY THAT THE SECOND CARD
IS A FACE CARD
GIVEN THAT THE FIRST CARD IS A FACE CARD?
AGAIN THERE ARE 12 FACE CARDS IN A DECK OF 52 CARDS.
SO IF WE KNOW THE FIRST CARD IS ALREADY A FACE CARD,
THEN THERE'S ONLY 51 CARDS LEFT IN THE DECK
AND IF ONE OF THE 12 FACE CARDS HAS ALREADY BEEN DRAWN,
THAT LEAVES 11 FAVORABLE OUTCOMES
AND THEREFORE THE PROBABILITY WOULD BE 11/51.
SO EVEN THOUGH THERE IS A SPECIAL FORMULA HERE
THAT CAN BE USED FOR CONDITIONAL PROBABILITY,
IF WE CAN MAKE SENSE OF THE SITUATION,
WE MAY NOT ACTUALLY HAVE TO USE IT.
I HOPE YOU FOUND THIS VIDEO HELPFUL,
THANK YOU FOR WATCHING.