A school survey shows that 10% of students are in band, 12% of students are in athletics, and 6% of students a

1. A school survey shows that 10% of students are in band, 12% of students are in athletics, and 6% of students are in both band and athletics. What is the probability that a student is in band or athletics?





2. Three coins are tossed. Find the probability that exactly 2 coins show heads if the first coin shows heads.





3. Given the integers 1 through 33, what is the probability that one of these integers is divisible by 4 if it is a multiple of 6.

A school survey shows that 10% of students are in band, 12% of students are in athletics, and 6% of students a
For any two events A and B the following is true





P(A or B) = P(A) + P(B) - P(A and B), or in more mathy terms:


P(A U B) = P(A) + P(B) - P(A ∩ B)





Let B be the event of being in the band


Let A be the event of being in athletics





Find P(A U B)





= P(A) + P(B) - P(A ∩ B)


= 0.12 + 0.10 - 0.06


= 0.16


























2)





we know from the binomial distribution that the unconditional probabilities for the number of heads showing in three tosses of a fair coin are:








P( X = 0 ) = 1/8


P( X = 1 ) = 3/8


P( X = 2 ) = 3/8


P( X = 3 ) = 1/8








for any two events A and B, with P(B) ≠ 0, then the conditional probability of A given B is:





P(A | B) = P( A ∩ B) / P(B)








We are looking for:





P( 2 heads | first toss is a head )


= P(2 heads and 1st is a head) / P(first is a head)


= (2/8) / (1/2)


= 0.5


























3)





multiples of 6 are:





{6, 12, 18, 24, 30}





2 of these are divisible by 4,





the probability is 2/5
Reply:1.





Take the number of students to be 100%.





The number of people in a band is 10%.





The number of people in athlethics is 12%





The number of people in both is 6%.





So 4% of people are only in a band (10%-6%)





And 6% are only in athletics (12% - 6%)








The possiblity of being in at least one is 4% + 6% + 6%





(People just in a band + Athletics peeps + peeps that do both)





=16%











2. Just because the first coin is a head does not impact the other two probabilities.





Eg. If i toss a coin and it is a head, the probability of me tossing and getting a head the next time is still 50/50.





So we are told the first coin is a heads and this is a definite and we can ignore it.





There are two coins left. The prob(heads) is 50% or 0.5





To get this with two coins we multiply





prob(heads) X prob(heads)





= 0.5 X 0.5 = 0.25 or 1/4





3.





The multiples of 6 between 1 - 33 are





6 12 18 24 30





12 and 24 are divisible by 4.





So the probability is 2 out of 5 = 2/5 = 0.4

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