Trending Content
| +81 | GF doesn’t care about her looks | 49 | 14h | |
| +55 | Affair with my Associate… In Desperate Need for Advice | 23 | 6h | |
| +41 | How to sound more eloquent? | 16 | 2h | |
| +39 | Insurance as High Finance? | 10 | 2d | |
| +38 | 400k/year in HVAC sales? | 26 | 1d | |
| +33 | Would you rather live alone in an outer borough or with roommates in Manhattan? | 23 | 2d | |
| +27 | Why do people listen to Jim Cramer Investing Advice? | 10 | 9h | |
| +26 | Is my boss gaslighting me? | 3 | 8h | |
| +25 | Carnivore Diet | 9 | 10h | |
| +25 | WSO Ranking On Resume??? | 6 | 6h |
Career Resources
Guess I'm wrong BOYS
Probability is 1/7 actually, as there are only 7 possible outcomes with 1 outcome being all 3 heads
rip
This is the answer I went with. 1/7
My other train of thought was that I already have 1 Heads. I only need 2 more from the remaining 2 coin flips, Thus (1/2)*(1/2) for 1/4
It's 1/7
Definitely 1/7
8 possibilities in total if you flip 3 coins. Knowing that there is at least one heads only rules out the possibility of having Tails Tails Tails, leaving 7 possibilities, one of which is three heads.
I got 1/7. For those of you interested in the detailed calculation: Let P(A) = Probability of getting 3 heads Let P(B) = Probability of getting at least 1 head or conversely 1 - probability of getting no heads. (1-(1/2)^3)
P(A) = (1/2)^3 = 1/8 P(B) = (3c1)(1/2)^3 + (3c2)(1/2)^3 + (3c3)(1/2)^3 = 7/8 by binomial theorem
Bayes: P(A|B) = P(B|A)*P(A)/P(B) P(B|A) = 1 (given you have 3 heads, you have a 100% chance of getting at least 1 head P(A) = 1/8 P(B) = 7/8
P(A|B) = P(B|A)*P(A)/P(B) = 1 * (1/8) / (7/8) = 1/7 Boom.
Strong username/post content ratio
FTW
Nice work
deleted
The correct answer is 0%. Your strong attraction to all forms of currency would cause you to grab the coin before it lands, investing it in a diverse portfolio which generates a 20% IRR.
Post removed for erroneous thinking.
25% If you know one of then is heads, then it leaves only 4 possible outcomes
H-HT H-HH H-TT H-TH
lol u srs?
Yeah except my thinking was flawed. Only one of the coins is Heads, not just the first one
Question requires a clarifying question - "at what point" which tells you whether or not to include the first heads.
More importantly these are independent events.
We did 3 flips, let's call them flip 1, filp 2, and flip 3.
There are two possible interpretations of 'you got at least one heads' I think of:
The interrogator decided to tell me the result of one of the flips. It hapened to be flip x which hapens to be heads. -> Two remaining flips need to be heads in order to have 3 heads, both of them are independent of x, which mean we have a probability of 1/4 of having 3 heads: 1/2 for the first remaining flip multiplied by 1/2 for the second one.
The interrogator tells me 'you got at least one heads' as a genuine description of what he is seeing, so that I know which are the possible states. -> For each flip there are 2 outcomes, meaning there is a total of 8 (2 x 2 x 2) outcomes. The assertion of the interogator only means that one of outcomes is not possible (flip 1 is not head, flip 2 is not head, flip 3 is not head), there are 7 remaining possible outcomes and only one the 7 is (flip 1 is head, flip 2 is head, flip 3 is head). The probability of having 3 heads is thus 1/7.
If you want the interrogator to be clear you can have the following discussion with him: - Which flip are you telling me about? - The second one. (for example, it doesn't really matter). - Would you had told me about the outcome of the second flip even if it was not heads?
Yes -> 1/4 No -> 1/7
.
OK, we are given that one coin was heads. If we knew that the given head was the first toss, then the odds of all three being heads are 1/4, as explained above. If we knew that the given head was the second toss, then the odds of all three being heads are 1/4. If we knew that the given head was the third toss, then the odds of all three being heads are 1/4. Sure, we do not know which toss is the given head, but it does not matter. Therefore, if we know that one coin is heads, then the odds that the other two are heads are still 1/4 (1/2 x 1/2).
This is simply incorrect. You are over thinking it. There are 8 equiprobable sets of 3 coin flips, 7 of which have at least 1 H, and only 1 of the 7 is HHH. See ActuarialQuant's post for math details. It is essentially the same trick as the famous boy/girl problem.
I stand corrected, though am still trying to correct my original thought process, which still makes sense, though apparently wrong. Thanks.
1/2
IlliniProgrammer HELP!
You tell the fuckers that the chances are 50-50 each time but your chances of closing a deal and getting all your work done is 100%. You're welcome
deleted
Ut in qui eos ea consequuntur. Ut vero rerum omnis omnis ut quidem. Sit reprehenderit quo molestiae dicta temporibus et cupiditate. Enim necessitatibus incidunt voluptas ea laboriosam eos esse. Voluptatem enim dolores quaerat nihil veritatis repellat non. Aut odio libero vero. Sint ratione quia porro.
Mollitia ea delectus rem nostrum. Accusamus odio esse pariatur adipisci. Est qui dolorem tempora voluptas deleniti sint.
See All Comments - 100% Free
WSO depends on everyone being able to pitch in when they know something. Unlock with your email and get bonus: 6 financial modeling lessons free ($199 value)
or Unlock with your social account...
Asperiores nisi maiores autem commodi. Eligendi doloremque praesentium eaque. Quo temporibus eaque asperiores excepturi eligendi impedit ipsa. Aut explicabo totam sed rerum ut repellendus hic perspiciatis. Voluptatem saepe dolores sed et harum est eos eligendi.