Question

QUESTION 1

Kids

Cars	1	2	3	4	5	6	Total
1	1	1	2	2	2	2	10
2	4	2	1	3	3	2	15
3	0	2	2	0	0	1	5
Total	5	5	5	5	5	5	30

Use the two-way table above to answer the following question.


Find the P(>3kids or ≥2cars).

 

QUESTION 2

Kids

Cars	1	2	3	4	5	6	Total
1	1	1	2	2	2	2	10
2	4	2	1	3	3	2	15
3	0	2	2	0	0	1	5
Total	5	5	5	5	5	5	30

Use the two-way table above to answer the following question.


Find the P(>1car | 3-5kids).

Answer 1

Solution:
We are given that following two-way table:

Question 1) Find the P(>3kids or ≥2cars).

Let A = Event that having > 3 kids and B = Event that having ≥2cars

So using an addition rule of probability, we get:

Thus find:

P(A) = P( having > 3 kids)

P(A) = P( 4 kids) + P(5 kids) + P( 6 kids)

From the above table, we can see the total of 4 kids is 5, the total of 5 kids is 5 and the total of 6 kids is also 5 and the grand total is N =30.

Thus

P(A) = (5/30) + (5/30) + (5/30)

P(A) = (5+5+5) / 30

P(A) = 15/30

P(A) = 0.5

Now find P(B):

P(B) = P( having ≥2cars)

P(B) = P( 2 cars ) + P(3 cars)

From above table, we can see total of 2 cars is 15 , total of 3 cars is 5

Thus

P(B) =15/30 + 5 /30

P(B) = (15+5)/30

P(B) = 20 / 30

P(B) =0.67

Now find :

From above table, numbers in Pink color represents outcomes which are having > 3 kids as well as having ≥2cars.

Thus

P( 4 kids , 2 cars) + P( 4 kids , 3 cars ) +

                        P( 5 kids , 2 cars ) + P(5 kids , 3 cars ) +

                 P( 6 kids , 2 cars ) + P(6 kids , 3 cars )

(3/30)+(3/30)+ (2/30)+P(0/30) + (0/30)+(1/30)

(3+3+2+0+0+1) / 30

9/30

0.3

Thus

Question 2) Find the P(>1car | 3-5kids).

Let C = > 1 Car and D = 3-5 kids

Thus

P( C | D) = ......?

Where

P( Having > 1 car as well as having 3-5 kids)

Outcomes in Yellow color represent having cars > 1 and having kids 3 to 5.

Thus

P(2 cars , 3 kids) + P(2 cars , 4 kids) + P(2 cars , 5 kids) +

                        P(3 cars , 3 kids) + P(3 cars , 4 kids) + P(3 cars , 5 kids)

(1/30) + (3/30)+(3/30)+ (2/30)+P(0/30) + (0/30)

(1+3+3+2+0+0)/30

9/30

0.3

now find P(D):

P(D) = P( having kids 3-5)

P(D) = P( 3 kids) + P(4 kids) + P(5 kids)

P(D) = (5/30)+(5/30)+(5/30)

P(D) = (5+5+5)/30

P(D) = 15/30

P(D) =0.5

Thus