Ej. Esperanza de vida
“X” va creciendo y “Y” va disminuyendo 1-X
Calidad de vida → se puede modificar
1.- A survey of groups viewing habitsover the last year revelaled the following information
28% watched gymnastic
29% watched basebal
19% watched soccer
14% watched gymnastic and baseball
12% watched baseball and soccer
8%watched all the three sports
Calculate the percentage of the group that watched none of the three sports during the last year.
A= watched gymnastic
B= watched baseball
C= watched soccer〖P(AυBυC)〗^C = 1- P(AυBυC)
1-(P(A) +P(B) + P(C)- (P(A P(A∩B) + P(A∩C) + P(B∩C)) + P(A∩B∩C))
= 1- (28% + 29% + 19% - (14% + 12% + 10%) + 8% = 52%
P(AυBυC)= P(AυB) + P(C) – P(AυBυ)∩C)
=P(A) + P(B) – P(A∩B) + P(C) – P((A∩C) υ (B∩C))
= (P(A) + P(B) + P(C) – (P(A∩B) +P(A∩C) + (B∩C)) + P(A∩B∩C))
2.- You are given P(AυB) = 007 And P(AυB’)= 0.9 Determine P(A):
P (AυB)=P(A) + P(B)- P(A∩B) = 0.7
P (AυB’)= P(A´) + P(B’) - P(A∩B’)= 0.9
P(AυB) + P(AυB) = 2 P(A) + 2 P(B’) - P(A∩B) - P(A∩B)
1.6= 0.7+0.9= 2 P(A) +1 – P(A)
= P (A) +1
P(A)= 0.63.- An urn contains 10 balls: 4 red and 6 blue, a second urn contains 16 red balls and unknown number of balls. A single ball is drawn from each urn. The probability that both balls are the same coloris 0.44. calculate the number of blue balls in the second urn.
4 red + 6 blue= 10 balls
P(R1)= 40 %
Ri= Event red ball I from urn i
Bi= Event blue ball I from urn i
0.44= P (R1 ∩R2) + P( B1 ∩ B2)
=P (R1) P (R2) + P(B1) + P (B2)
=(4/16) ( 16/x+16) + (6/16) (x/16) = (64/10x+160) + (6x/10x+160) = 64+6x/10x +160
A public healthresearcher the medical records of a group of 937 men who died in 1998 and discovers that 210 of the men died from causes related to heart disease.
More over 312 of the 937 men had at least one parent who...
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