Answer:
[tex]P(X\cap Y\cap Z)=0.05[/tex]
Step-by-step explanation:
From the question we are told that:
Percentage of audience in first concert [tex]P(X)=0.28[/tex]
Percentage of audience in second concert [tex]P(Y)=0.42[/tex]
Percentage of audience in third concert [tex]P(Z)=0.30[/tex]
Audience Percentage at at-least one concert [tex]P(X \cup Y \cup Z)=0.80[/tex]
Percentage of audience at first & second concert [tex]P(X \cap Y)=0.10[/tex]
Percentage of audience in first & third concert [tex]P(X \cap Z)=0.08[/tex]
Percentage of audience in second & third concert [tex]P(Y\cap Z)=0.07[/tex]
Generally the equation for probability of attending all concerts [tex]P(X\cap Y\cap Z)[/tex]is mathematically given by
[tex]P(X \cup Y \cup Z)=P(X)+P(Y)+P(Z)-P(X \cap Y)-P(X \cap Z)-P(Y\cap Z)+P(X\cap Y\cap Z)[/tex]
[tex]P(X\cap Y\cap Z)=P(X \cup Y \cup Z)-P(X)-P(Y)-P(Z)+P(X \cap Y)+P(X \cap Z)+P(Y\cap Z)[/tex]
[tex]P(X\cap Y\cap Z)=0.80-0.28-0.42-0.30+0.10+0.80+0.70[/tex]
[tex]P(X\cap Y\cap Z)=0.05[/tex]
Therefore the probability that a randomly selected audient attended all the concerts
[tex]P(X\cap Y\cap Z)=0.05[/tex]
