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Re: If a certain coin is flipped, the probability that the coin [#permalink]
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26 Feb 2012, 23:53
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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 [#permalink]
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03 Apr 2012, 21:43
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 [#permalink]
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04 Apr 2012, 00:39
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 [#permalink]
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10 Jul 2013, 09:54
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 [#permalink]
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06 Mar 2016, 10:30
Bunuel wrote: 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. In this problem is the above the equivalent of some of these other answers (and how I approached the problem) of simply doing Chance for tails & chance for heads both = 1/2 Chance of heads 1/2 * chance of heads *chance of heads 1/2 * chance of heads 1/2 * chance of tails 1/2 * chance of tails 1/2? I see the above total as 2^5 over 1 possibility so the numbers are the same but for straight forward problems such as this one is the listed 1/2*1/2*1/2 an acceptable approach or just a coincidence it works for this problem?
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Re: If a certain coin is flipped, the probability that the coin will land [#permalink]
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06 Jan 2017, 23:07
This question is simple since the coin is ideal(i.e., 0.5 probability for either event) The question asks us to find the probability of the occurrence of the event HHHTT This is just the product, P(H)*P(H)*P(H)*P(T)*P(T)=(0.5)*(0.5)*(0.5)*(0.5)*(0.5) =1/32 If the coin was skewed,i.e., say probability of landing head is 0.75,then the answer would have been, P(H)*P(H)*P(H)*P(T)*P(T)=(0.75)*(0.75)*(0.75)*(0.25)*(0.25) =27/1024
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Re: If a certain coin is flipped, the probability that the coin [#permalink]
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06 Feb 2017, 09:38
The probability of getting heads is 1/2 Similarly, the probability of getting tails is also 1/2 The probability of getting heads on the first 3 flips is (1/2)^3 = 1/8 Similarly, the probability of getting tails on the last 2 flips is (1/2)^2 = 1/4 So, the required probability is 1/8*1/4 = 1/32
Answer : E



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Re: If a certain coin is flipped, the probability that the coin [#permalink]
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10 Jun 2017, 09:07
Bunuel wrote: 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. hey, could we also calculate it as follows: (1/2)^5 * (5*4*3)/(3*2*1)
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Re: If a certain coin is flipped, the probability that the coin [#permalink]
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10 Jun 2017, 09:16
daviddaviddavid wrote: Bunuel wrote: 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. hey, could we also calculate it as follows: (1/2)^5 * (5*4*3)/(3*2*1)Thew correct answer is 1/2^5, so the answer is no. What is the logic behind what you wrote?
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Re: If a certain coin is flipped, the probability that the coin [#permalink]
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10 Jun 2017, 09:44
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 [#permalink]
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Re: If a certain coin is flipped, the probability that the coin [#permalink]
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10 Jun 2017, 09:50
Bunuel wrote: daviddaviddavid wrote: 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 =) Yes, that's correct. thank you very much =)
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