According to a survey of 100 people regarding their movie attendance in the last year, 40 had seen a science fiction movie, 55 had seen an adventure movie, and 20 had seen a horror movie. Moreover, 25 had seen a science fiction movie and an adventure movie, 5 had seen an adventure movie and a horror movie, and 15 had seen a science fiction movie and a horror movie. Only 5 people had seen a movie from all three categories.

The probability that at least one movie had seen P(S∪(AD)∪H) = 0.75

Step-by-step explanation:

The total number of movie attendance in the last year n =100

Let 'S' be the event had seen science fiction movie

The probability of that the event had seen science fiction movie

P(S) = 40/100 = 0.4

Let 'AD' be the event had seen adventure  movie

the probability of that  the event had seen adventure  movie

P(AD) =55/100 = 0.55

Let 'H' be the event had seen Horror movie

The probability of that the event had seen Horror movie

P(H) =20/100 = 0.2

given data 25 had seen a science fiction movie and an adventure movie

The probability of that the seen a science fiction movie and an adventure movie

that is P(S∩(AD)) = 25/100=0.25

Given  5 had seen an adventure movie and a horror movie

The probability of that the seen adventure movie and a horror movie

That is P((AD)∩ H) = 5/100=0.05

Given  5 had seen an science movie and a horror movie

The probability of that the seen science movie and a horror movie

That is P((S∩ H) = 15/100=0.15

Given only 5 people had seen a movie from all three categories.

The probability of that the seen a movie from all three categories.

That is P(S∩(AD)∩H) =5/100=0.05

The probability that at least one movie had seen P(S∪(AD)∪H) =

              =  P(S)+P(AD)+P(H) - P(S∩(AD)-P((AD)∩ H)-P((S∩ H) + P(S∩(AD)∩H)

              = 0.4+0.55+0.2-0.25-0.05-0.15+0.05

              = 0.75

the answer is 0.000652422

but if you ment to ask what is 6131 divided by 4, it is 1532.75.

step-by-step explanation: