Mathematics, 25.05.2020 19:59, drew3vans34

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)

Answers: 2

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answered: bks53

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!

answered: Guest

child does not get an education so she can use the molla on junk

answered: Guest

First, think of what you know. you know that the number t the end of the equation will need to be subtracted because it is the money he lost from buying the lawn mower. and your first two weeks have negative profit, so that tells you that c and d are wrong. now you can plug any of the weeks into a and b to see which one works. i'll start with plugging week 1 into a. $y = 60(1) - 90 \\ y = 60 - 90 \\ y = - 30$ this isn't right because it says your answer for week one should be -90.so now we need to try the same thing with b. $y = 60(1) - 150 \\ y = 60 - 150 \\ y = - 90$ this one got you the right answer, (sometimes if you only check with one of the weeks multiple will work so you need to check with more of the weeks, but it worked out fine here)this means that b is the correct answer.let me know if you need more clarification. i hope this : )