Question

12. An undercover news reporter wanted to know what exactly Fast Food hamburgers were made of. It was found that 95% of the burgers were made from beef, the rest was made from a Mystery Meat!

a) What is the probability that at least 18 out of 20 of hamburgers selected randomly are made of beef? [4]

Answer 1

I tried #12. I'm not sure if it's right though.
@jim_thompson5910

Answer 2

This is another binomial distribution problem.


P(X = x) = (n C x)*(p)^(x)*(1-p)^(n-x)

P(X = 18) = (20 C 18)*(0.95)^(18)*(1-0.95)^(20-18)

P(X = 18) = (20 C 18)*(0.95)^(18)*(0.05)^(20-18)

P(X = 18) = (190)*(0.95)^(18)*(0.05)^2

P(X = 18) = (190)*(0.397214318)*(0.0025)

P(X = 18) = 0.18867680105


P(X = x) = (n C x)*(p)^(x)*(1-p)^(n-x)

P(X = 19) = (20 C 19)*(0.95)^(19)*(1-0.95)^(20-19)

P(X = 19) = (20 C 19)*(0.95)^(19)*(0.05)^(20-19)

P(X = 19) = (20)*(0.95)^(19)*(0.05)^1

P(X = 19) = (20)*(0.3773536025)*(0.05)

P(X = 19) = 0.3773536025


P(X = x) = (n C x)*(p)^(x)*(1-p)^(n-x)

P(X = 20) = (20 C 20)*(0.95)^(20)*(1-0.95)^(20-20)

P(X = 20) = (20 C 20)*(0.95)^(20)*(0.05)^(20-20)

P(X = 20) = (1)*(0.95)^(20)*(0.05)^0

P(X = 20) = (1)*(0.3584859224085)*(1)

P(X = 20) = 0.3584859224085


=====================================================================

P(X >= 18) = P(X = 18) + P(X = 19) + P(X = 20)


P(X >= 18) = 0.18867680105 + 0.3773536025 + 0.3584859224085


P(X >= 18) = 0.9245163259585


So the probability that at least 18 out of 20 of hamburgers selected randomly are made of beef is 0.9245163259585

Answer 3

Oh, nice! I had the numbers I just didn't know what to do with them. Thank you!

Answer 4

sure thing

Answer 5

Hey, I sent an attachment. Would it be possible if you checked two other questions out? Would you prefer if I opened the questions in separate convos?

Answer 6

@jim_thompson5910

Answer 7

What did you get for the answer of #1?

Answer 8

I got binomial.

Answer 9

for which part?

Answer 10

a)

Answer 11

Is replacement allowed?

Answer 12

Is replacement another form of distribution?

Answer 13

no I meant, are the cards replaced or put back when you draw them?

Answer 14

It isn't mentioned, so I would guess no.

Answer 15

so then a hypergeometric distribution is used

Answer 16

a binomial distribution is only used if the probability of each trial is the same

Answer 17

Oh, so the next one is binomial then.

Answer 18

yes that's what I'd go with too

Answer 19

Actually, I think the answer is uniform.

Answer 20

This is a uniform probability distribution since there can only be one outcome and the probability of rolling a single die is the same.

Answer 21

Ah very true, wasn't thinking that

Answer 22

If it were binomial, there would be two possible outcomes, success or failure.

Answer 23

yes very true, I was thinking of rolling a 3 or something

Answer 24

Why do you think a) is hypergeometric? I'm a bit perplexed.

Answer 25

Because I'm reading that the hypergeometric distribution is like the binomial distribution, but replacements aren't made (so the probabilities will change)

Answer 26

This is what I read---> However, while the binomial distribution relies on independent trials, the hypergeometric distribution relies on dependent trials. The probability of success changes with each successive trial.

Answer 27

that's basically what I'm reading as well

Answer 28

so the second draw depends on the first draw because the cards aren't replaced

Answer 29

Hey so, is the last one binomial since the coin is re-tossed?

Answer 30

hmm don't see one about a coin flip

Answer 31

Hey, finding the probability of drawing a certain colour of candy out of a box of candies would be an example of a binomial. This is similar since the question is looking for how many fours are rolled.

Answer 32

I'm not sure what you mean by similar question

Answer 33

I mean the question is similar to the example. The example mentions that finding the probability of a CERTAIN colour of candy whereas in the question (1c.), it's about rolling a CERTAIN number, which in this case is 4.

Answer 34

ah i gotcha

Answer 35

well if you defined success as landing on that color, then you could argue that it is a binomial distribution

Answer 36

Alright.

Answer 37

Hey, did you open the attachment I sent? It includes a graph. Not sure how I could sent it otherwise on this thread, lol.

Answer 38

I don't see a graph in it

Answer 39

Oh, sorry. I mean chart, lol.

Answer 40

oh yes, i see it

Answer 41

where did you need help with on this one?

Answer 42

Well, I tried doing #6 but I'm not sure if I was doing it right.

Answer 43

I did this at first.

n= 2+ 4+6+8+10+12/ 6= 7
p= 0.12+0.31+0.04+0.18+0.22+0.13/ 6 = 0.17

E(x)= np
= 7 (0.17)
= 1.19

Answer 44

Then, I tried this.

E(x)= 2(0.12) + 4 (0.31) + 6 (0.04) + 8(0.18) + 10(0.22) + 12(0.13)
= 6.92

Answer 45

@jim_thompson5910

Answer 46

The answer of 6.92 is the correct answer, so your second answer and work for it is correct.

Answer 47

Thanks I submitted my work for this question using my first attempt to my teacher earlier and she replied that "your numbers are correct, but what you did with them is not".

Answer 48

@jim_thompson5910

Answer 49

I see, I'm glad you got some of it correct.

Answer 50

Thanks, but now I'm a bit skeptical since the solution that I for the first attempt was 1.19.

Answer 51

that I got*

Answer 52

The solution is still 6.92

The numbers in your first attempt are correct, it's just how you used them is incorrect which gave you the wrong answer.

Answer 53

OOOOOH. Ok, lol.