A fair coin is tossed three times and the events AA, BB, and CC are defined as follows: A:{A:{ At least one head is observed }} B:{B:{ At least two heads are observed }} C:{C:{ The number of heads observed is odd }} Find the following probabilities by summing the probabilities of the appropriate sample points (note that 0 is an even number):

This question is incomplete, the complete question is;

A fair coin is tossed three times and the events AA, BB, and CC are defined as follows: A:{A:{ At least one head is observed }} B:{B:{ At least two heads are observed }} C:{C:{ The number of heads observed is odd }} Find the following probabilities by summing the probabilities of the appropriate sample points (note that 0 is an even number): a) P(B) =, b) P(A or B) =

a) probability of B is 0.5

b) probability of (A or B)  is 0.875

Step-by-step explanation:

Given the data in the question;

The coin is tossed 3 times.

so let S represent the sample space with the probabilities;

S = [ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT ]

so n(S) = 8

now let A be the set where at least one head is observed;

A = [ HHH, HHT, HTH, THH, HTT, THT, TTH, ]

n( A) = 7

let B be the set where at least two head is observed;

B = [ HHH, HHT, HTH, THH, ]

n(B) = 4

let C be the set where number of heads are odd;

C = [ HHH, HTT, THT, TTH ]

n(C) = 4

so

a) P(B) = ?

probability of B will be;

P(B) = n(B) / n(S)

we substitute

P(B) = 4 / 8 = 0.5

Therefore, probability of B is 0.5

b) P(A or B) = ?

the favorable number of cases to event A∪B is

A∪B = [ HHH, HHT, HTH, THH, HTT, THT, TTH, ]∪[ HHH, HHT, HTH, THH, ]  

A∪B = [ HHH, HHT, HTH, THH, HTT, THT, TTH, ]

n(A∪B) = 7

Now the probability of A or B will be

P( A∪B ) = n(A∪B) / n(S) = 7/8 = 0.875

Therefore, probability of (A or B)  is 0.875

$20.01

step-by-step explanation:

the sale price is (1.00 - 0.77)($87), or 0.23($87), or $20.01.

so, converting firstly the mixed fractions to improper fractions and then "adding".