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.

Answer 1

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`