Respuesta :
Answer:
a) We need to check two conditions:
1) [tex] \sum_{i=1}^n P_i = 1[/tex]
[tex] 0.05+0.14+0.34+0.24+0.11+0.07+0.02+0.02+0.01= 1[/tex]
2) [tex]P_i \geq 0 , \forall i=1,2,...,n[/tex]
So we satisfy the two conditions so then we have a probability distribution
b) [tex] P(C \geq 1)[/tex]
And we can use the complement rule and we got:
[tex] P(C \geq 1)= 1-P(C<1) = 1-P(C=0)=1-0.05=0.95[/tex]
c) [tex] P(C=0) = 0.05[/tex]
d) For this case we see that the result from part b use the probability calculated from part c using the complement rule.
Step-by-step explanation:
For this case we have the following probability distribution given:
C 0 1 2 3 4 5 6 7 8
P 0.05 0.14 0.34 0.24 0.11 0.07 0.02 0.02 0.01
And we assume the following questions:
a) Verify that this is a probability distribution
We need to check two conditions:
1) [tex] \sum_{i=1}^n P_i = 1[/tex]
[tex] 0.05+0.14+0.34+0.24+0.11+0.07+0.02+0.02+0.01= 1[/tex]
2) [tex]P_i \geq 0 , \forall i=1,2,...,n[/tex]
So we satisfy the two conditions so then we have a probability distribution
b) What is the probability one randonmly chosen classmate has at least one child
For this case we want this probability:
[tex] P(C \geq 1)[/tex]
And we can use the complement rule and we got:
[tex] P(C \geq 1)= 1-P(C<1) = 1-P(C=0)=1-0.05=0.95[/tex]
c) What is the probability one randonmly chosen classmate has no children
For this case we want this probability:
[tex] P(C=0) = 0.05[/tex]
d) Look at the answers for parts b and c and explain their relationship
For this case we see that the result from part b use the probability calculated from part c using the complement rule.
