Experimental Probability
A farmer asks, ‘Will it rain this month?’. The answer to the farmer’s question depends on three things; the months, the place where the farmer is, and what has happened in the past three months in that place. The table below gives some answers to the question for different places and months.
Place | Month | Answer to question |
Sokoto | February | No |
Jos | July | Ye |
Ibadan | January | Maybe |
Port Harcourt | June | yes |
Is it possible to give a more accurate answer to a farmer near Ibadan in January? The table below shows that on average, 10mm of rain falls in Ibadan in January. However, this is an average found by keeping records over twelve years. The actual rainfall for Ibadan in January over the 12 years was as follows.
J | F | M | A | M | J | J | A | S | O | N | D | |
Sokoto | 0 | 0 | 0 | 10 | 48 | 91 | 155 | 249 | 145 | 15 | 15 | 0 |
Jos | 3 | 3 | 28 | 56 | 203 | 226 | 330 | 292 | 213 | 41 | 3 | 3 |
Ibadan | 10 | 23 | 89 | 137 | 150 | 188 | 160 | 84 | 178 | 155 | 46 | 10 |
Port Harcourt | 66 | 109 | 155 | 262 | 404 | 660 | 531 | 318 | 516 | 460 | 213 | 81 |
18 mm | 0 mm | 17 mm | 9 mm |
11 mm | 22 mm | 14 mm | 0 mm |
16 mm | 0 mm | 7 mm | 6 mm |
From the above data, it can be seen that rain fell in nine of the twelve months on January. If future years follow the pattern of the past, it is likely that in Ibadan, rain will fall in nine out of the next 12 Januaries. We say that theprobability of rain falling in Ibadan in January is 9/12 (or ¾ or 0.75 or 75%). This probability can never be exact. However, it is the best measure that we can give from the data we have. The number 9/12 is based on the experimental records. We say that it is experimental probability.
Example 1
A girl writes down the number of male and female children of her mother and father. She also writes down the number of male and female children of the parents’ brothers and sisters. Her resuts are shown in table below.
Number of male children | Number of female children | |
Mother and father | 2 | 5 |
Mother’s brother | 6 | 8 |
Mother’s sister | 4 | 8 |
Father’s brother | 5 | 8 |
Father’s sister | 7 | 7 |
Total | 24 | 36 |
Solution
In the girl’s family there is a total of 60 children. 36 of these are female. If the girl’s own children follow the pattern of her family, then the experimental probability that her first born will be a girl is 36/6 = 3/5.
Number of male children that the girl is likely to have = 2/5 of 5 =2.
Notice that the results in the above example are based on experimental probability. Thus we are using the past to predict the future. Events can easily turn out differently. The answers in the example above are no more that calculated guesses.
Exercise
Probability as a fraction
Probability is a measure of the likelihood of a required outcome happening. It is usually as a fraction:
Probability = number of required outcomes/number of possible outcomes
In Example above, the required outcomes were female children and the possible outcomes were both male and female children. Thus probability of having a female child
= number of female children/number of male and female children
= 36/60 = 3/5 = 0.6
If we are completely sure that something will happen, the probability is 1. For example, if today is Tuesday, the probability that tomorrow is Wednesday is 1.
If we are sure that something cannot happen, the probability is 0. For example, the probability of rolling a 7 on a pencil is 0, because there is no number 7 in the pencil. If the probability of something happening is x, then the probability of it not happening is 1 – x. For example, if the probability of it raining month is 9/12, then the probability of it not raining is 3/12.
Example
Number of cars developing faults = 50
Number of cars altogether = 1 000
Probability of buying a faulty car = 50/1 000 = 1/20
‘At random means without carefully choosing’.
Either:
Four out of 100 oranges are bad, thus 96 out of 100 oranges are good.
Probability of getting a good orange = 96/100 = 24/25
or:
probability of getting a bad orange = 4/100 = 1/25
thus,
probability of getting a good orange = 1 – 1/25 = 24/25.
The next example shows how to use probability when analyzing statistical data.
Example
City school enters candidates for the WASSCE. The results for the years 2004 – 2008 are given in the table below:
Year | 2004 | 2005 | 2006 | 2007 | 2008 |
Number of candidates | 86 | 93 | 102 | 117 | 116 |
Number gaining WASSCE passes | 51 | 56 | 57 | 65 | 70 |
Solution
= 51 + 56 + 57 + 65 + 70
= 299
Total number of candidates
= 86 + 93 + 102 + 117 + 116
= 514
Success rate as a percentage
= 299/514
=0.58 to 2.s.f.
Success rate as a percentage = 0.58 X 100% = 58%
In part b it is assumed that the student’s chances of success are the same as the school’s success rate.
Exercise
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