A jar contains 4 purple, 3 blue, and 1 gold ticket. Three tickets are selected without replacement.
Find the probability that:
a: all of the tickets are purple
b: all of the tickets are blue
c: the first two are purple and the third is gold
d: none of the tickets is blue.
(please include explanation if possible, tysm)

a) 1/14

b) 1/56

c) 1/28

d) 5/28

Step-by-step explanation:

a) Since tickets are being selected without replacement, this means that the pool of outcomes diminishes with each selection. For the first pick, there are a total of 8 tickets to choose from, and 4 of those are purple. Therefore, the probability that the first ticket is purple is 4/8=1/2. For the second ticket, the total number of remaining tickets is 7, and the number of purple tickets is 3, meaning that the probability is 3/7. For the final ticket, 6 remaining tickets and 2 of them being purple yields a probability of 2/6=1/3. Multiplying all of these together, you get 1/2*3/7*1/3=1/14.

b) Similarly to the previous problem, the first probability is 3/8, followed by 2/7, and then by 1/6. Multiplying these together, you get a probability of 1/56.

c) The probability that the first two tickets are purple are 1/2 and 3/7, as we have already calculated. For the final ticket, there are 6 possible remaining with 1 of them being gold. This means a probability of 1/6, which yields a total of 1/28.

d) Since there are 5 tickets originally that aren't blue, the probability for the first pick is 5/8, followed by 4/7 and 3/6=1/2. Multiplying these together, you get a final answer of 5/8*4/7*1/2=5/28.

Hope this helps!

(x+3)(x+7)

step-by-step explanation: