## (New Version Available) Conditional Probability

– 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
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 RED FACE CARD GIVEN THAT THE FIRST CARD
IS A RED 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.