Probability from Experimental Data
The (empirical) probability of an event \(E\) is the number of trials in which \(E\) occurs divided by the
total number of trials.
Example 16.1 Two coins are tossed 500 times, and we get two heads 105 times, one head 275 times and no head 120 times.
Find the probability of each event.
Solution. Let \(E_1\), \(E_2\), \(E_3\) be the events two heads, one head and no head. Then
\[
P(E_1)=\frac{105}{500}=0.21,\qquad P(E_2)=\frac{275}{500}=0.55,\qquad P(E_3)=\frac{120}{500}=0.24.
\]
The probabilities of all the elementary events add up to \(1\):
\[
P(E_1)+P(E_2)+P(E_3)=0.21+0.55+0.24=1.
\]
Example 16.2 A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced.
The table shows the results of 1000 cases.
lcccc@
Distance (km) | Less than 4000 | 4000 to 9000 | 9001 to 14000 | More than 14000 |
Frequency | 20 | 210 | 325 | 445 |
If a tyre is bought from this company, what is the probability that
(i) it has to be replaced before 4000 km is covered;
(ii) it will last more than 9000 km;
(iii) it has to be replaced after it has covered between 4000 km and 14000 km?
Solution. (i) Out of \(1000\) trials, \(20\) tyres were replaced before 4000 km:
\[
P(E_1)=\frac{20}{1000}=0.02.
\]
(ii) Tyres lasting more than 9000 km: \(325+445=770\), so
\[
P(E_2)=\frac{770}{1000}=0.77.
\]
(iii) Tyres replaced between 4000 km and 14000 km: \(210+325=535\), so
\[
P(E_3)=\frac{535}{1000}=0.535.
\]
Example 16.3 The percentages of marks obtained by a student in the monthly tests are given below.
lccccc@
Test | 1 | 2 | 3 | 4 | 5 |
Percentage of marks obtained | 69 | 71 | 73 | 68 | 74 |
Based on this table, find the probability of the student getting more than 70
Solution. Of the \(5\) tests, \(3\) (71, 73, 74) have more than 70
\[
P(\text{scoring more than 70\% marks})=\frac35=0.6.
\]
