## Basic Probability, part 2

### August 23, 2009

Another way to consider the probability is graphical way.

Consider the following case:
there are a coin and the five balls in a bag with numbers 1,2,3,4,5 respectively. We toss a coin and then take out from the bag a ball.
Let us draw first event – tossing the coin, there can be H with the probability $\frac{1}{2}$ and T with the probability $\frac{1}{2}$:

Pretend that we get H, then we take a ball with the probability $\frac{1}{5}$ for each number. Notice that the occasion there will be the same if we get T, and then take a ball with the probability $\frac{1}{5}$ :

As we know to calculate the probability of two events we need to multiply them. Thus the probability of the event “T on the coin and the ball number 1” is $P(E)=\frac{1}{2}* \frac{1}{5}=\frac{1}{10}$.
The probability of the event “T on the coin and the ball number 2” is $P(E)=\frac{1}{2}* \frac{1}{5}=\frac{1}{10}$.
The probability of the event “H on the coin and the ball number 3” is $P(E)=\frac{1}{2}* \frac{1}{5}=\frac{1}{10}$, and so on.

And as we know the probability of all cases is equal 1:
the probability of getting H or T on the coin is $P(E)=\frac{1}{2}+ \frac{1}{2}=1$;
the probability of taking the ball with the number 1 or 2 or 3 or 4 or 5 5 is $P(E)=\frac{1}{5}+ \frac{1}{5}+ \frac{1}{5}+ \frac{1}{5}+ \frac{1}{5}=1$;
the probability of getting H or T on the coin and then taking the ball with the number 1 or 2 or 3 or 4 or 5 is $P(E)=\frac{1}{10}+ \frac{1}{10}+ \frac{1}{10}+ \frac{1}{10}+ \frac{1}{10}+ \frac{1}{10}+ \frac{1}{10}+ \frac{1}{10}+ \frac{1}{10}+ \frac{1}{10}=1$. You can see it on the previous picture.

Score a goal.

Get an another example. We are watching football game and we know that the forward scored 300 goals in 1000 attacks. So the property that he score a goal in an attack is $P(E)=\frac{3}{10}.$ If there were 2 attacks, what is the probability that the forward scored the goal on the first attack and missed on the second attack. To get answer we need to calculate the probability of the missing in an attack, it is $P(Miss)=1-P(Score)=1- \frac{3}{10}= \frac{7}{10},$ since the forward could miss or score and nothing else. Denote “Score a goal” as S, “Miss a goal” as M, and show all possible variants on the picture:

As before we multiplied the probabilities. The probability that the forward scored the goal on the first attack and missed on the second attack is $P(SM)=\frac{21}{100}.$

For 3 or more events, for instance, when 3 attacks occurred we need to draw a picture with a larger tree (scheme). The idea is the same.

Score a goal in the five attempts.

Consider the case when we play the basketball and have 5 attempts to throw the ball. If we know that we can get a score with the probability $P(S)=\frac{2}{3},$ what is the probability that we at least take one score in all 5 attempts. What does it mean? It means that we need to find all possible variants of the scoring except one – when we miss every time in all five attempts. In other words, we find $P(E)=1 - P(MMMMM),$ and it will be the answer.
To calculate $P(MMMMM)$ let us find the probability of missing in an attempt, it is $P(M)=1-P(S)=1-\frac{2}{3}=\frac{1}{3}.$
Hence the probability that we lose all attempts is $P(MMMMM)=P(M)*P(M)*P(M)*P(M)*P(M)=\\ \frac{1}{3}*\frac{1}{3}*\frac{1}{3}*\frac{1}{3}*\frac{1}{3}=\frac{1}{243}.$

Then the probability that we at least take one score in all 5 attempts is $1-\frac{1}{243}=\frac{242}{243}.$ And it is very high, so probably we take a score.