Question

An organization selected 2400 families at random and survey them to determine a relationship between income level and the number of vehicle in a family. The information gathered is listed in the table below:

Monthly income (Rs.)	Vehicles per family
	0	1	2	Above 2
Less than 7,000	10	160	25	0
7,000-10,000	0	305	27	2
10,000-13,000	1	535	29	1
13,000-16,000	2	469	59	25
16,000 or more	1	579	82	88

Suppose a family is chosen. Find the probability that the family chosen is
(i) Earning Rs.10,000-13,000 per month and owning exactly 2 vehicles.
(ii) Earning Rs.16,000 or more per month and owning exactly 1 vehicle.
(iii) Earning less than Rs.7,000 per month and does not own any vehicle.
(iv) Earning Rs.13,000-16,000 per month and owning more than 2 vehicles.
(v) Owning not more than 1 vehicle.

Answer 1

Hint: To find out the probability of a desired outcome we are provided with the formulae
P(A) =n(E)/n(S)
Where,
P(A) = Probability of an event
n(E) = Number of desired outcome
n(S) = Total number of outcomes


Complete step by step solution:
Case (i): Earning Rs.10,000-13,000 per month and owning exactly 2 vehicles.
1. Total number of families = Total number of outcomes
i.e. n(S) = 2400
2. Number of families earning Rs. 10,000-13,000 per month and owning exactly 2 vehicles = Number of desired outcomes
i.e. n(E) = 29
3. If P(A) to be the probability of choosing a family that earns Rs. 10,000-13,000 per month and owns exactly 2 vehicles
Using the formulae P(A) = n(E)/n(S)
Substituting the values of n(E) and n(S) from step 1 and step 2, we get
P(A) = $\dfrac{{29}}{{2400}}$
P(A) Probability of choosing a family that earns Rs. 10,000-13,000 per month and owns exactly 2 vehicles = $\dfrac{{29}}{{2400}}$
Case (ii): Earning Rs.16,000 or more per month and owning exactly 1 vehicle.
4. Total number of families = Total number of outcomes
i.e. n(S) = 2400
5. Number of families earning Rs. 16,000 or more per month and owning exactly 1 vehicle = Number of desired outcomes
i.e. n(E) = 579
6. If P(A) to be the probability of choosing a family that earns Rs. 10,000-13,000 per month and owns exactly 2 vehicles
Using the formulae P(A) = n(E)/n(S)
Substituting the values of n(E) and n(S) from step 4 and step 5, we get
P(A) = $\dfrac{{579}}{{2400}}$
P(A) Probability of choosing a family that earns Rs. 16,000 or more per month and owns exactly 1 vehicle = $\dfrac{{579}}{{2400}}$
Case (iii): Earning less than Rs.7,000 per month and does not own any vehicle.
7. Total number of families = Total number of outcomes
i.e. n(S) = 2400
8. Number of families earning less than Rs.7,000 per month and does not own any vehicle. = Number of desired outcomes
i.e. n(E) = 10
9. If P(A) to be the probability of choosing a family that Earns less than Rs.7,000 per month and does not own any vehicle.
Using the formulae P(A) = n(E)/n(S)
Substituting the values of n(E) and n(S) from step 7 and step 8, we get
P(A) = $\dfrac{{10}}{{2400}}$
P(A) Probability of choosing a family that earns less than Rs.7,000 per month and does not own any vehicle = $\dfrac{{10}}{{2400}}$
Case (iv): Earning Rs.13,000-16,000 per month and owning more than 2 vehicles.
10. Total number of families = Total number of outcomes
i.e. n(S) = 2400
11. Number of families earning Rs.13,000-16,000 per month and owning more than 2 vehicles = Number of desired outcomes
i.e. n(E) = 25
12. If P(A) to be the probability of choosing a family that Earns Rs.13,000-16,000 per month and owning more than 2 vehicles
Using the formulae P(A) = n(E)/n(S)
Substituting the values of n(E) and n(S) from step 10 and step 11, we get
P(A) = $\dfrac{{25}}{{2400}}$
P(A) Probability of choosing a family that earns less than Rs.7,000 per month and does not own any vehicle = $\dfrac{{25}}{{2400}}$
Case (v): Owning not more than 1 vehicle.
13. Total number of families = Total number of outcomes
i.e. n(S) = 2400
14. Number of families owning not more than 1 vehicles = Number of desired outcomes
i.e. n(E) = families having monthly income less than Rs. 7,000 and having 0 and 1 vehicle + families having monthly income Rs. 7,000-10,000 and having 0 and 1 vehicle + families having monthly income Rs. 10,000-13,000 and having 0 and 1 vehicle + families having monthly income Rs. 13,000-16,000 and having 0 and 1 vehicle + families having monthly income Rs. 16,000 or more and having 0 and 1 vehicle
= 10+160+0+305+1+535+2+469+1+579
= 2,062
15. If P(A) to be the probability of choosing a family owning not more than 1 vehicle
Using the formulae P(A) = n(E)/n(S)
Substituting the values of n(E) and n(S) from step 13 and step 14, we get
P(A) = $\dfrac{{2062}}{{2400}}$
P(A) Probability of choosing a family owning not more than 1 vehicle = $\dfrac{{2062}}{{2400}}$


Note: It should be taken into consideration that if a student wants to check whether the above probabilities in each case is right or not then add all the above probabilities. If the sum of all the probabilities comes out to be one then the above probabilities are right because the sum of all the probabilities is always equal to one.