Question

A city hosted a music festival that included three concerts. According to the sales database, 28% of the audience attended the first concert, 42% attended the second one, 30% attended the third one, 80% attended at least one of the three concerts, 10% attended the first and second ones, 8% attended the first and third ones, and 7% attended the second and third ones. Find the probability that a randomly selected audient attended all the concerts

Answer 1

Answer:

$P(X\cap Y\cap Z)=0.05$

Step-by-step explanation:

From the question we are told that:

Percentage of audience in first concert $P(X)=0.28$

Percentage of audience in second concert $P(Y)=0.42$

Percentage of audience in third concert $P(Z)=0.30$

Audience Percentage at at-least one concert $P(X \cup Y \cup Z)=0.80$

Percentage of audience at first & second concert $P(X \cap Y)=0.10$

Percentage of audience in first & third concert $P(X \cap Z)=0.08$

Percentage of audience in second & third concert $P(Y\cap Z)=0.07$

 

Generally the equation for probability of attending all concerts $P(X\cap Y\cap Z)$is mathematically given by

$P(X \cup Y \cup Z)=P(X)+P(Y)+P(Z)-P(X \cap Y)-P(X \cap Z)-P(Y\cap Z)+P(X\cap Y\cap Z)$

$P(X\cap Y\cap Z)=P(X \cup Y \cup Z)-P(X)-P(Y)-P(Z)+P(X \cap Y)+P(X \cap Z)+P(Y\cap Z)$

$P(X\cap Y\cap Z)=0.80-0.28-0.42-0.30+0.10+0.80+0.70$

$P(X\cap Y\cap Z)=0.05$

Therefore the probability that a randomly selected audient attended all the concerts

$P(X\cap Y\cap Z)=0.05$