Conditional probabilityRoger Federer and Raphael Nadal are playing the final of Wimbledon's tennis tournament. If Federer wins the first set then he has 8 chances in 10 to win the match. If Nadal wins the first set then he has one chance in 2 to win the match. Nadal has only 3 chances in 10 to win the first set, what is the probability that he wins the match ?
There are two options, he can win the match, having previously either won the first set or lost it. You must add the two probabilities. To have a better view of it, let's call S the event 'Nadal wins the first set' and M the event 'Nadal wins the match', and let's draw it.
, the probability of M given S, is the probability that Nadal wins the match knowing that he won the first set. According to this statement, the probability is ˝, you write it down on the tree. you write down on the tree all the probabilities that you can calculate.
In the same way, the event 'Nadal wins the first set then wins the match' is the event . Its probability is equal to .
Finally, the probability Nadal wins the match is equal to the sum of the two probabilities we have just calculated, it is the formula of total probabilities .
Raphael Nadal has a little less than 3 chances in 10 to win the match.
The ones who are not in terminale S finished the lesson about probabilities, The ones who are in S can go on.
DénombremementA box contains 5 balls numbered from 1 to 5, the first 3 ones are green, the last 2 ones are red.
1. You pick up a ball. What is the probability that it is green? It's easy, and it's for starters:
2. You pick up consequently 5 balls and put them back in the box after you pick up each of them. What is the probability that you draw 5 red balls? The answer is
Probabilities are multiplying.
3. You pick up consequently 3 balls without putting them again in the box. What is the probability that you get 2 green ones and a red one? You must examine all cases which lead to the result and add their probabilities. You can pick up VVR, VRV ou RVV hence the probability is :
4. How many different ways are there to draw simultaneously 3 balls in the box (for example draw the balls 123, 125, 345, but not 132 because it was counted already)?
The answer is number . It's the number of subsets of 3 elements contained in a set of 5 elements. You calculate it with the formula :
Number n! reads 'factorial n'.
There are 10 different ways.
5. You draw 3 balls simultaneously. What is the probability that you draw 2 green ones and one red (event E)? The number of favorable cases is equal to the number of possibilities to draw 2 green ones among 3 green ones multiplied by the number of possibilities to draw one red ball among the 2 red ones. The total number of cases is 10 (question 4). The detailed calculation is as follows :
Binomial theoremThe numbers are also useful to develop an math of the kind . Indeed :