Each of fifty students participating in a raffle is assigned a different number from 1 to 50 to write on his/her ticket. if a single number is picked as a winner, answer the following questions.
a) what is the probability of an odd-numbered ticket winning, given that, without exceeding 50, all of the students who were assigned an odd number, multiplied it by 2?
b) what is the probability of a an even-numbered ticket winning, given that the students who were initially assigned a prime number, all added 3 to their number?
c)what is the probability of an even-numbered ticket winning, given that, all of the student who were assigned a consecutive neighbor of a two-digit prime number added 1 to their number?

a) 0.24

b) 0.76

c) 0.14

Step-by-step explanation:

Between 1 and 50, there are 25 odd numbers and 25 even numbers.

a) The first 13 odd numbers (1 to 25) are doubled, becoming even numbers.  The last 12 odd numbers (27 to 49) are left.

P(odd) = 12/50

P(odd) = 0.24

b) There are 15 prime numbers less than 50 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47).  They are all odd except for 2.  Adding 3 to each number, 1 prime number changes from even to odd, and 14 prime numbers change from odd to even.  There are now 38 even numbers.

P(even) = 38/50

P(even) = 0.76

c) Prime numbers are odd.  Numbers consecutive to prime numbers are even.  Adding 1 to those numbers, they all become odd.  So the only even numbers left are those that are not 2-digit numbers and those not consecutive to a prime number.  That means there are 7 even numbers left (2, 4, 6, 8, 26, 34, and 50).

P(even) = 7/50

P(even) = 0.14

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)