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Question
A company selected 2400 families at random and survey them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:
| Monthly income: (in Rs) |
Vehicles per family | |||
| 0 | 1 | 2 | Above 2 | |
| Less than 7000 7000-10000 10000-13000 13000-16000 16000 or more |
10 0 1 2 1 |
160 305 535 469 579 |
25 27 29 29 82 |
0 2 1 25 88 |
If a family is chosen, find the probability that family is:
(i) earning Rs10000-13000 per month and owning exactly 2 vehicles.
(ii) earning Rs 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs 7000 per month and does not own any vehicle.
(iv) earning Rs 13000-16000 per month and owning more than 2 vehicle.
(v) owning not more than 1 vehicle
(vi) owning at least one vehicle.
Solution
The total number of trials is 2400.
Remember the empirical or experimental or observed frequency approach to probability.
If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted byP (A) and is given by
` P(A) = m/n`
(i) Let A1 be the event that a chosen family earns Rs 10000-13000 per month and owns exactly 2 vehicles.
The number of times A1 happens is 29.
Therefore, we have` P (A_1) = 29/2400`
(ii) Let A2 be the event that a chosen family earns Rs 16000 or more per month and owns exactly 1 vehicle.
The number of times A2 happens is 579.
Therefore, we have ` P (A_2) = 579/2400`
(iii) Let A3 be the event that a chosen family earns less than Rs 7000 per month and does not owns any vehicles.
The number of times A3 happens is 10.
Therefore, we have
` P (A_3) = 10/2400`
=`1/240`
(iv) Let A4 be the event that a chosen family earns Rs 13000-16000 per month and owns more than 2 vehicles.
The number of times A4 happens is 25.
Therefore, we have
` P (A_4) = 25/2400`
=`1/96`
(v) Let A5 be the event that a chosen family owns not more than 1 vehicle (may be 0 or 1). In this case the number of vehicles is independent of the income of the family.
The number of times A5 happens is
(10+0+1+2+1)+(160+305+535+469+579)=2062.
Therefore, we have
` P (A_5) = 2062/2400`
=`1031/1200`
(vi) Let A6 be the event that a chosen family owns atleast 1 vehicle (may be 1 or 2 or above 2). In this case the number of vehicles is independent of the income of the family.
The number of times A6 happens is
(160+305+535+469+579)+(25+27+29+29+82)+(0+2+1+25+88)=2356 .
Therefore, we have
` P (A_6) = 2356/2400`
=`589/600`
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