Probability question - getting the proa

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discover1234

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Hello,

First time posting here! I've been really stumped on this particular math problem for my math class, and I'm hoping you guys might be able to help me out.

Problem below: I've already done a) but I'm stuck on b). I'm pretty sure I got a) right, but just in case it's wrong, I'll mention that I got 158, 940, 114, 100, 040 as my answer using the combinations formula, nCr = n! / r! * (n - r)!.

a) gives me the sample size for 11 coins to be selected from 101 regardless of whether the selected coin is fake or not, but I can't figure out the way to use this information and which formula to get the answer to b). I know that to find the probability for something that is "at least one" is by finding the compliment of the event (so probability of no fake coins selected) and subtracting it from 1, but I'm not sure where to go from here. I'd really appreciate some help guys, thank you!

There are 101 coins in a box, three of which are fake and 98 are genuine. Eleven randomly selected coins are tested.

a) In how many ways can eleven coins be selected from the 101 coins in the box? Give the answer in factorial notation.

b) What is the probability that at least one of the fake coins is selected?
 
Note that part (a) asks for the answer not as a number, but in factorial notation. But your answer is of course correct.

For part (b), use the standard "trick" (as you suggested) that P(at least one A) = P(not all non-A) = 1 - P(all non-A). So, what is the probability that none of the fake coins is selected? (That is, that all of the selected coins are genuine.)
 
Consider two groups--fake and real.
How many ways can you pick the number of fakes coins you want out of the number of fakes coins?
How many ways can you pick the number of real coins you want out of the number of real coins.
How many coins in total do you want to pick out of the total number of coins?

Your answer will involve the three numbers above. How should they be arranged?

Think hypergeometric!
 
Hey guys, thanks for getting back to me. This is what I got, if you could let me know if I'm right or at least on the right track that'd be great.

(E) = number of combinations to get at least one fake coin from the 11 randomly selected of the 101
(not E) = number of combinations for all 98 real coins (excluded from this pool is the 3 fake ones) from the 11 randomly selected coins
P(E) = 1 - P(not E) and P(not E) = 1 - P(E)

Combinations formula...for (not E). nCr = n!/r!* (n-r)! -> n = 98, r = 11 -> 98!/11!(87)! -> 112, 044, 912, 118, 048 (this answer is (not E)), as this is the chance of getting 1 fake coin to all fake coins, represented by every combination that can do this.

1 in the formula is equal to my number answer for a), which represents every combination, real or fake coins included, for 11 randomly drawn coins out of the 101. That gave me 158, 940, 114, 100, 040

P(E) = 1 - P(not E)...so 158, 940, 114, 100, 040 - 112, 044, 912, 118, 048 = 46, 895, 201, 981, 992. This gives me (E).

And now to show the probability as a fraction....

46, 895, 201, 981, 992
------------------------
158, 940, 114, 100, 040

simplified....

149
----
505

Is this the correct answer? Thanks!
 
discover1234 said:
Problem below: I've already done a) but I'm stuck on b). I'm pretty sure I got a) right, but just in case it's wrong, I'll mention that I got 158, 940, 114, 100, 040 as my answer using the combinations formula, nCr = n! / r! * (n - r)!.
There are 101 coins in a box, three of which are fake and 98 are genuine. Eleven randomly selected coins are tested.
a) In how many ways can eleven coins be selected from the 101 coins in the box? Give the answer in factorial notation.
b) What is the probability that at least one of the fake coins is selected?
Click to expand...
I am answering this to stress notation. Suppose that \(\bf{E}\) denotes the event of selecting eleven off those coins.
Then \(\#(\bf{E})=\dfrac{101!}{11!\cdot 90!}\) is the number of ways to have \(\bf{E}\).
Now suppose \(\bf{N}\) is the event that in selecting eleven of the coins none of the fake coins is included.
Do you follow that \(\#(\bf{N})=\dfrac{98!}{11!\cdot 87!}~?\)
Does it follow that \(\dfrac{\#(\bf{N})}{\#(\bf{E})}\) is the probability that none of the chosen eleven is fake?
Now can you finish and post the result?
 
Last edited:
My answer is further up, I have to wait for moderators to approve my posts, so just in case someone misses my initial answer it's there (the 4th post).

b) didn't ask for notation (I corrected my a) to 101! / 11! (90!) as the actual answer, rather than the number version) so I did it with numbers, but I suppose it makes sense to have that one done via factorial notation too.

So answering in factorial notation instead...

P(E) = 1 - P(not E) and P(not E) = 1 - P(E)

1 = 101! / 11! (90)!
P(not E) = 98!/11!(87)!

101! / 11! (90)!- 98!/11!(87)! = (E)

P(E) =
(101! / 11! (90)!- 98!/11!(87)!)
-------------------------------
101! / 11! (90)!

So that answer would be better than my answer in numbers in post 4 of this thread? Is how I have written P(E) OK, or should it give one factorial notation as an answer? I know how to calculate the number result from the subtraction there, but putting it in a final factorial form I am unsure of.

Thanks for your help.
 
Good work!

It didn't explicitly ask for factorial notation in part (b), so my sense is that a decimal answer (0.295) would be accepted; but your simplified fraction 149/505 is excellent, and if they did want factorials, your answer at the end of post #6 is also good.

Different authors use different notation for the number of outcomes in an event, so I don't know whether your teacher may use the notation you used, like "(E)"; but pka's notation of #(E) is clearer; I've seen others use n(E). In the end, though, you clearly meant the right things, though what you said in some places was wrong if taken literally.

Again, good work.
 
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