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If a certain coin is flipped, the probability that the coin
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If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips? A. 3/5 B. 1/2 C. 1/5 D. 1/8 E. 1/32
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Originally posted by marcodonzelli on 12 Jan 2008, 02:44.
Last edited by Bunuel on 25 Feb 2012, 22:34, edited 1 time in total.
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Re: If a certain coin is flipped, the probability that the coin
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25 Feb 2012, 22:51
marcodonzelli wrote: If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips? A. 3/5 B. 1/2 C. 1/5 D. 1/8 E. 1/32 We want the probability that coin will land heads up on the first 3 flips and not on the last 2 flips. So there is ONLY one favorable outcome, namely heads up on the first 3 flips and tails up on the last 2 flips: HHHTT. # of total out comes is 2^5=32. P=favorable/total=1/32. Answer: E. IF THE QUESTION WERE: A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails) Then the answer would be: \(P=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}\), since in this case the winning scenario of HHHTT can occur in many ways: HHHTTT, HTHHT, HTTHH, ... which basically is # of permutations of 5 letters out of which 3 H's and 2 T's are identical, so \(\frac{5!}{3!2!}\). As you can see the probability of the event for modified question is 10 times higher that the probability of the event for the initial question, and that's because # of favorable outcomes is 10 for modified question and 1 for initial question. Hope it's clear.
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Re: coins
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12 Jan 2008, 05:23
Its like this:
On the first three flips, you must get heads. Whats the probability of getting heads ? Its 1/2
So for the first three flips, your probability is (1/2)^3 = 1/8
Now for the last two, you want to get tails only. Whats the prob of getting tails ? Well, its the same as prob of getting a heads, namely, 1/2
For the last two flips, your probability is (1/2)^2 = 1/4
So your overall probability for the event in question is 1/8*1/4 = 1/32




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Re: coins
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12 Jan 2008, 18:18
The probability of landing heads and not landing on heads is same = ½ The probability of first three heads = ½* ½ * ½ The probability of last two landing not on heads = ½ * ½ The total probability = ½ * ½ * ½ * ½ * ½ = 1/ 32 Answer: E



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Re: coins
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27 Sep 2009, 05:57
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?
A. 3/5 B. 1/2 C. 1/5 D. 1/8 E. 1/32
Soln: Ans is E: 1/32
Total number of outcomes is = 2 * 2 * 2 * 2 * 2 = 32 ways
Now it says that first 3 heads and 2 tails {HHHTT} => of the above 32 possible outcomes, in only one outcome will we get {HHHTT}
THus probability is = 1/32



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Re: coins
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01 Aug 2011, 08:28
Bernoulli can't be used here  it will give you the probability of getting 3 heads or 2 tail, not the probability of getting 3 heads followed by two tails. Bernoulli doesn't take order into account, only successes (however you define it) within a certain number of trials. I suppose a misapplication of the Bernoulli formula will yield the correct answer, but it's important to understand how the formula really works if you're going to use it.
Personally, I've never seen a probability question where knowing the Bernoulli formula would give one a significant advantage... I suspect that a lot of time is wasted memorizing such formulas.
The simplest way to think about this is that HHHTT is just one outcome among 32 possible outcomes.



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Re: coins
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01 Aug 2011, 12:47
The coin tosses are mutually exclusive.. result of the first coin toss doesnt affect the result of second coin toss..
pr. for heads on first toss = 1/ 2 pr. for heads on second toss = 1/ 2 pr. for heads on third toss = 1/ 2 pr. for tails on 4th toss = 1/ 2 pr. for tails on 5th toss = 1/ 2
= (1/2)^ 5 = 1/ 32
if it was a bag of red balls and blue balls.. then drawing for a red ball would have affected the probability for the next draw.. here is not the case..



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Re: If a certain coin is flipped, the probability that the coin
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25 Feb 2012, 18:52
When I see a coin or dice question, I automatically tend to think using:  the technique that khan academy references in his tutorial on probabilities (that's the technique that came to my mind first in this case)  the binomial formula b(n;x,p)  which walker tends to use in a few instances here.
In this case: 5 trials, each trial has 2 possibilities; total outcomes = 2^5 = 32 The possibility we're looking for is a unique outcome: HHHTT  which can only occur in one way. So, probability = 1/32
Another technique that the khan academy references: P (H and H and H and T and T) = P(H) * P(H) * P(H) * P(T) * P(T) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32
I strongly recommend watching the khan academy videos if the sight of coin & dice questions makes you quiver, like it did for me. (From a GMAT Verbal standpoint, is the above sentence grammatically correct? should it be "make" or "makes"  and i'm struggling with the last part "like it did for me"...)



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Re: If a certain coin is flipped, the probability that the coin
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26 Feb 2012, 04:30
Bunuel wrote: IF THE QUESTION WERE: A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)
Then the answer would be: \(P=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}\), since in this case the winning scenario of HHHTT can occur in many ways: HHHTTT, HTHHT, HTTHH, ... which basically is # of permutations of 5 letters out of which 3 H's and 2 T's are identical, so \(\frac{5!}{3!2!}\).
As you can see the probability of the event for modified question is 10 times higher that the probability of the event for the initial question, and that's because # of favorable outcomes is 10 for modified question and 1 for initial question.
Hope it's clear. in this case we can also use the following formula nCk/2^n = 5C3/2^5=10/32
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Re: If a certain coin is flipped, the probability that the coin
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26 Feb 2012, 04:44
LalaB wrote: Bunuel wrote: IF THE QUESTION WERE: A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)
Then the answer would be: \(P=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}\), since in this case the winning scenario of HHHTT can occur in many ways: HHHTTT, HTHHT, HTTHH, ... which basically is # of permutations of 5 letters out of which 3 H's and 2 T's are identical, so \(\frac{5!}{3!2!}\).
As you can see the probability of the event for modified question is 10 times higher that the probability of the event for the initial question, and that's because # of favorable outcomes is 10 for modified question and 1 for initial question.
Hope it's clear. in this case we can also use the following formula nCk/2^n = 5C3/2^5=10/32 Yes, you can: both give exactly the same expression, since \(C^3_5=\frac{5!}{3!2!}\). Though your reasoning for 5C3 could be different from what I used in my approach: selecting which 3 flips out of 5 will give heads up.
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Re: If a certain coin is flipped, the probability that the coin
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26 Feb 2012, 21:25
Although for bunuel's extension of the original question, seems like he was using the binomial prob formula...m trials, n successes, prob of success p, prob of failure q ( or 1p) B = mCn * p^n * q^(mn) ......with m=5, n=3, p=q=1/2



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Re: If a certain coin is flipped, the probability that the coin
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26 Feb 2012, 22:53
fortsill wrote: Although for bunuel's extension of the original question, seems like he was using the binomial prob formula...m trials, n successes, prob of success p, prob of failure q ( or 1p) B = mCn * p^n * q^(mn) ......with m=5, n=3, p=q=1/2 Yes, it's a formula for binomial distribution explained here: mathprobability87244.htmlSome questions to practice: whatistheprobabilitythatafieldgunwillhitthriceon127334.htmlthereisa90chancethataregisteredvoterinburghtown56812.htmlcombinationps55071.htmltheprobabilitythatafamilywith6childrenhasexactly88945.html? Hope it helps.
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Re: If a certain coin is flipped, the probability that the coin
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03 Apr 2012, 20:43
whats the difference between this question and the following?
If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?
is it because it could any 3 days between the 4th and 8th?



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Re: If a certain coin is flipped, the probability that the coin
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03 Apr 2012, 23:39
dchow23 wrote: whats the difference between this question and the following?
If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?
is it because it could any 3 days between the 4th and 8th? The difference is explained here: ifacertaincoinisflippedtheprobabilitythatthecoin58357.html#p1049993The probability of rain each day is 1/2 and the probability of no rain is also 1/2. \(C^3_5=10\) represent ways to choose on which 3 days out of 5 there will be a rain, so \(P=C^3_5*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}=\frac{5}{16}\). Or think about it this way: we want the probability of the following event: RRRNN, where R represent rain day and N represents norain day. Now, each R and each N have individual probability of 1/2, so \((\frac{1}{2})^5\). But the case of RRRNN can occur in many ways: RRRNN, RRNRRN, RNRRN, NRRRN, ... basically it will be equla to # of arrangements (permutations) of 5 letters RRRNN out of which there are 3 identical R's and 2 identical N's. That # of arrangements is \(\frac{5!}{3!2!}\), (notice that it's the same as \(C^3_5\)). So, finally \(P=\frac{5!}{3!2!}*(\frac{1}{2})^5=\frac{5}{16}\). For more on this topic check Combinations and Probability chapters of Math Book: mathcombinatorics87345.htmlmathprobability87244.htmlAlso check similar questions to practice: iftheprobabilityofrainonanygivendayis50whatis99577.htmlonsaturdaymorningmalachiwillbeginacampingvacation100297.htmlwhatistheprobabilitythatafieldgunwillhitthriceon127334.htmlthereisa90chancethataregisteredvoterinburghtown56812.htmlcombinationps55071.htmltheprobabilitythatafamilywith6childrenhasexactly88945.html? Hope it helps.
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Re: If a certain coin is flipped, the probability that the coin
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10 Jul 2013, 08:54
My try ( I am a beginner trying to solve this kind of problems using combinations \(\frac{C^5_3}{{(C^2_1)^5}}\) = 1/32 \(C^5_3\) : number of combinations in which we obtain 3 H in 5 tosses \((C^2_1)^5\) : number of combinations of T or H in 5 tosses
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Re: If a certain coin is flipped, the probability that the coin
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10 Jun 2017, 08:44
hey, could we also calculate it as follows: (1/2)^5 * (5*4*3)/(3*2*1)[/quote] Thew correct answer is 1/2^5, so the answer is no. What is the logic behind what you wrote?[/quote] hey thanks for you answer I meant to your modified question: (A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails) (1/2)^5 > probability * (5*4*3)/(3*2*1) > combinations many thanks =)
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Re: If a certain coin is flipped, the probability that the coin
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Re: If a certain coin is flipped, the probability that the coin
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06 Feb 2018, 10:50
Hi All, This prompt is actually a remarkably straightforward probability question, since it asks you for an EXACT series of results based on coin flips: HHHTT Since there's a 1/2 chance of flipping heads and a 1/2 chance of flipping tails, the probability of flipping that exact sequence of events is: (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If a certain coin is flipped, the probability that the coin
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23 Apr 2018, 12:11
marcodonzelli wrote: If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?
A. 3/5 B. 1/2 C. 1/5 D. 1/8 E. 1/32 We want P(heads on 1st AND heads on 2nd AND heads on 3rd AND tails on 4th AND tails on 5th) = P(heads on 1st) x P(heads on 2nd) x P(heads on 3rd) x P(tails on 4th) x P(tails on 5th) = 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32 Answer: E Cheers, Brent
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Re: If a certain coin is flipped, the probability that the coin
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26 Dec 2018, 05:14
marcodonzelli wrote: If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?
A. 3/5 B. 1/2 C. 1/5 D. 1/8 E. 1/32
\(? = P\left( {{\text{specific}}\,\,{\text{4}}\,\,{\text{landings}}\,\,{\text{sequence}}} \right) = \frac{1}{{32}}\) \({\rm{Reason}}:\,\,\,\left\{ \matrix{ \,{2^4} = 32\,\,\,{\rm{equiprobable}}\,\,{\rm{possible}}\,\,{\rm{outcomes}} \hfill \cr \,1\,\,\, = \,\,\,{\rm{specific}}\,\,{\rm{4}}\,\,{\rm{landings}}\,\,{\rm{sequence}}\,\,{\rm{desired}}\,\,\,({\rm{favorable)}} \hfill \cr} \right.\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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